Physics Practice MCQs
Units, Measurement, and Dimensional Analysis
1
The unit of thermal conductivity is
Correct Option: (a)
\(K=\dfrac{Qx}{A(T_1-T_2)t}\), where \(Q\) is the amount of heat flow, \(x\) is the thickness of the slab, \(A\) is the area of cross section, and \(t\) is the time taken.
\[ K=\frac{Jm}{m^2Ks} =W\frac{1}{m}\frac{1}{K} =Wm^{-1}K^{-1} \]
\(K=\dfrac{Qx}{A(T_1-T_2)t}\), where \(Q\) is the amount of heat flow, \(x\) is the thickness of the slab, \(A\) is the area of cross section, and \(t\) is the time taken.
\[ K=\frac{Jm}{m^2Ks} =W\frac{1}{m}\frac{1}{K} =Wm^{-1}K^{-1} \]
2
The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are
Correct Option: (c)
Damping force, \(F\propto v\) or \(F=kv\), where \(k\) is the constant of proportionality.
\[ k=\frac{F}{v} =\frac{N}{ms^{-1}} =\frac{kg\,m\,s^{-2}}{m\,s^{-1}} =kg\,s^{-1} \]
Damping force, \(F\propto v\) or \(F=kv\), where \(k\) is the constant of proportionality.
\[ k=\frac{F}{v} =\frac{N}{ms^{-1}} =\frac{kg\,m\,s^{-2}}{m\,s^{-1}} =kg\,s^{-1} \]
3
The unit of permittivity of free space, \( \varepsilon_0 \), is
Correct Option: (c)
Force between two charges, \[ F=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2} \] Therefore, \[ \varepsilon_0= \frac{1}{4\pi}\frac{q^2}{Fr^2} =\frac{C^2}{N\,m^2} \]
Force between two charges, \[ F=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2} \] Therefore, \[ \varepsilon_0= \frac{1}{4\pi}\frac{q^2}{Fr^2} =\frac{C^2}{N\,m^2} \]
4
A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is
Correct Option: (c)
Given : least count = 0.01 mm and number of circular scale divisions = 50.
\[ \text{Pitch} =\text{L.C.}\times \text{No. of circular scale divisions} \] \[ =0.01\times 50 =0.5\ \text{mm} \]
Given : least count = 0.01 mm and number of circular scale divisions = 50.
\[ \text{Pitch} =\text{L.C.}\times \text{No. of circular scale divisions} \] \[ =0.01\times 50 =0.5\ \text{mm} \]
5
In an experiment, the percentage of error occurred in the measurement of physical quantities \(A, B, C\) and \(D\) are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement \(X\), where
\[
X=\frac{A^2B^{1/2}}{C^{1/3}D^3}
\]
will be
Correct Option: (c)
\[ X=\frac{A^2B^{1/2}}{C^{1/3}D^3} \] Maximum percentage error in \(X\), \[ \frac{\Delta X}{X}\times100 = \left( 2\frac{\Delta A}{A} +\frac12\frac{\Delta B}{B} +\frac13\frac{\Delta C}{C} +3\frac{\Delta D}{D} \right)\times100 \] \[ =2\times1+\frac12\times2+\frac13\times3+3\times4 =16\% \]
\[ X=\frac{A^2B^{1/2}}{C^{1/3}D^3} \] Maximum percentage error in \(X\), \[ \frac{\Delta X}{X}\times100 = \left( 2\frac{\Delta A}{A} +\frac12\frac{\Delta B}{B} +\frac13\frac{\Delta C}{C} +3\frac{\Delta D}{D} \right)\times100 \] \[ =2\times1+\frac12\times2+\frac13\times3+3\times4 =16\% \]
6
The main scale of a vernier callipers has \(n\) divisions/cm. \(n\) divisions of the vernier scale coincide with \((n-1)\) divisions of main scale. The least count of the vernier callipers is
Correct Option: (c)
If \(n\) divisions of vernier scale coincide with \((n-1)\) divisions of main scale.
\[ nVSD=(n-1)MSD \] \[ 1VSD=\frac{n-1}{n}MSD \] \[ \text{Least Count}=1MSD-1VSD \] \[ =1MSD-\frac{n-1}{n}MSD \] \[ =\frac{1}{n}MSD \] Since, \[ 1MSD=\frac{1}{n}\;cm \] Therefore, \[ \text{Least Count} =\frac{1}{n}\times\frac{1}{n} =\frac{1}{n^2}\;cm \]
If \(n\) divisions of vernier scale coincide with \((n-1)\) divisions of main scale.
\[ nVSD=(n-1)MSD \] \[ 1VSD=\frac{n-1}{n}MSD \] \[ \text{Least Count}=1MSD-1VSD \] \[ =1MSD-\frac{n-1}{n}MSD \] \[ =\frac{1}{n}MSD \] Since, \[ 1MSD=\frac{1}{n}\;cm \] Therefore, \[ \text{Least Count} =\frac{1}{n}\times\frac{1}{n} =\frac{1}{n^2}\;cm \]
7
A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is
Correct Option: (d)
Least count \(=0.001\;cm\)
Main scale reading \(=5\;mm=0.5\;cm\)
Circular scale reading \(=25\) divisions
Observed reading \[ =0.5+25\times0.001 =0.525\;cm \] Zero error \[ =-0.004\;cm \] Correct diameter \[ =0.525-(-0.004) =0.529\;cm \]
Least count \(=0.001\;cm\)
Main scale reading \(=5\;mm=0.5\;cm\)
Circular scale reading \(=25\) divisions
Observed reading \[ =0.5+25\times0.001 =0.525\;cm \] Zero error \[ =-0.004\;cm \] Correct diameter \[ =0.525-(-0.004) =0.529\;cm \]
8
In an experiment, four quantities \(a, b, c\) and \(d\) are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity \(P\) is calculated as follows
\[
P=\frac{a^3b^2}{cd}
\]
% error in \(P\) is
Correct Option: (c)
\[ P=\frac{a^3b^2}{cd} \] Percentage error in \(P\), \[ \frac{\Delta P}{P}\times100 = \left[ 3\left(\frac{\Delta a}{a}\right) + 2\left(\frac{\Delta b}{b}\right) + \frac{\Delta c}{c} + \frac{\Delta d}{d} \right]\times100 \] \[ =3\times1\%+2\times2\%+3\%+4\% \] \[ =14\% \]
\[ P=\frac{a^3b^2}{cd} \] Percentage error in \(P\), \[ \frac{\Delta P}{P}\times100 = \left[ 3\left(\frac{\Delta a}{a}\right) + 2\left(\frac{\Delta b}{b}\right) + \frac{\Delta c}{c} + \frac{\Delta d}{d} \right]\times100 \] \[ =3\times1\%+2\times2\%+3\%+4\% \] \[ =14\% \]
9
A student measures the distance traversed in free fall of a body, initially at rest, in a given time. He uses this data to estimate \(g\), the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are \(e_1\) and \(e_2\) respectively, the percentage error in the estimation of \(g\) is
Correct Option: (b)
From the relation, \[ h=ut+\frac{1}{2}gt^2 \] Since body is initially at rest, \[ h=\frac{1}{2}gt^2 \] \[ g=\frac{2h}{t^2} \] Taking natural logarithm on both sides, \[ \ln g=\ln h-2\ln t \] Differentiating, \[ \frac{\Delta g}{g} = \frac{\Delta h}{h} - 2\frac{\Delta t}{t} \] For maximum permissible error, \[ \left(\frac{\Delta g}{g}\times100\right) = \left(\frac{\Delta h}{h}\times100\right) + 2\left(\frac{\Delta t}{t}\times100\right) \] Given, \[ \frac{\Delta h}{h}\times100=e_1, \qquad \frac{\Delta t}{t}\times100=e_2 \] Therefore, \[ \left(\frac{\Delta g}{g}\times100\right) =e_1+2e_2 \]
From the relation, \[ h=ut+\frac{1}{2}gt^2 \] Since body is initially at rest, \[ h=\frac{1}{2}gt^2 \] \[ g=\frac{2h}{t^2} \] Taking natural logarithm on both sides, \[ \ln g=\ln h-2\ln t \] Differentiating, \[ \frac{\Delta g}{g} = \frac{\Delta h}{h} - 2\frac{\Delta t}{t} \] For maximum permissible error, \[ \left(\frac{\Delta g}{g}\times100\right) = \left(\frac{\Delta h}{h}\times100\right) + 2\left(\frac{\Delta t}{t}\times100\right) \] Given, \[ \frac{\Delta h}{h}\times100=e_1, \qquad \frac{\Delta t}{t}\times100=e_2 \] Therefore, \[ \left(\frac{\Delta g}{g}\times100\right) =e_1+2e_2 \]
10
If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be
Correct Option: (d)
\[ V=\frac{4}{3}\pi R^3 \] Taking logarithm, \[ \ln V=\ln\left(\frac{4}{3}\pi\right)+3\ln R \] Differentiating, \[ \frac{dV}{V}=3\frac{dR}{R} \] Hence, \[ \frac{\Delta V}{V}\times100 =3\times\frac{\Delta R}{R}\times100 \] Given error in radius \(=2\%\), \[ \text{Error in volume} =3\times2\% =6\% \]
\[ V=\frac{4}{3}\pi R^3 \] Taking logarithm, \[ \ln V=\ln\left(\frac{4}{3}\pi\right)+3\ln R \] Differentiating, \[ \frac{dV}{V}=3\frac{dR}{R} \] Hence, \[ \frac{\Delta V}{V}\times100 =3\times\frac{\Delta R}{R}\times100 \] Given error in radius \(=2\%\), \[ \text{Error in volume} =3\times2\% =6\% \]
11
The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in the measurement of mass and lengths are 3% and 2% respectively, the maximum error in the measurement of density would be
Correct Option: (d)
Maximum error in mass \[ \frac{\Delta m}{m}=3\%=\frac{3}{100} \] Maximum error in length \[ \frac{\Delta l}{l}=2\%=\frac{2}{100} \] Since \[ \rho=\frac{m}{l^3} \] Therefore, \[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\frac{\Delta l}{l} \] \[ = \frac{3}{100} + 3\times\frac{2}{100} \] \[ = \frac{9}{100} =9\% \]
Maximum error in mass \[ \frac{\Delta m}{m}=3\%=\frac{3}{100} \] Maximum error in length \[ \frac{\Delta l}{l}=2\%=\frac{2}{100} \] Since \[ \rho=\frac{m}{l^3} \] Therefore, \[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\frac{\Delta l}{l} \] \[ = \frac{3}{100} + 3\times\frac{2}{100} \] \[ = \frac{9}{100} =9\% \]
12
Percentage errors in the measurement of mass and speed are 2% and 3% respectively. The error in the estimate of kinetic energy obtained by measuring mass and speed will be
Correct Option: (a)
Percentage error in mass \[ \frac{\Delta m}{m}=2\% \] Percentage error in speed \[ \frac{\Delta v}{v}=3\% \] Kinetic Energy, \[ K.E.=\frac{1}{2}mv^2 \] Therefore, \[ \frac{\Delta(K.E.)}{K.E.} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v} \] \[ =2\%+2(3\%) \] \[ =8\% \]
Percentage error in mass \[ \frac{\Delta m}{m}=2\% \] Percentage error in speed \[ \frac{\Delta v}{v}=3\% \] Kinetic Energy, \[ K.E.=\frac{1}{2}mv^2 \] Therefore, \[ \frac{\Delta(K.E.)}{K.E.} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v} \] \[ =2\%+2(3\%) \] \[ =8\% \]
13
A certain body weighs 22.42 g and has a measured volume of 4.7 cc. The possible error in the measurement of mass and volume are 0.01 g and 0.1 cc. Then maximum error in the density will be
Correct Option: (b)
Density, \[ \rho=\frac{m}{V} \] Taking logarithm, \[ \ln\rho=\ln m-\ln V \] Differentiating, \[ \frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} - \frac{\Delta V}{V} \] Errors are always added, \[ \frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V} \] \[ = \left( \frac{0.01}{22.42} + \frac{0.1}{4.7} \right)\times100 \] \[ \approx 2\% \]
Density, \[ \rho=\frac{m}{V} \] Taking logarithm, \[ \ln\rho=\ln m-\ln V \] Differentiating, \[ \frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} - \frac{\Delta V}{V} \] Errors are always added, \[ \frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V} \] \[ = \left( \frac{0.01}{22.42} + \frac{0.1}{4.7} \right)\times100 \] \[ \approx 2\% \]
14
Taking into account of the significant figures, what is the value of
\[
9.99\,m-0.0099\,m \, ?
\]
Correct Option: (b)
\[ 9.99-0.0099 =9.9801 \] In subtraction, the result should be reported according to the least precise decimal place. Since 9.99 has two decimal places, \[ 9.9801 \approx 9.98 \] Hence the answer is \[ 9.98\,m \]
\[ 9.99-0.0099 =9.9801 \] In subtraction, the result should be reported according to the least precise decimal place. Since 9.99 has two decimal places, \[ 9.9801 \approx 9.98 \] Hence the answer is \[ 9.98\,m \]
15
Dimensions of stress are
Correct Option: (d)
Stress is defined as \[ \text{Stress} = \frac{\text{Force}}{\text{Area}} \] Dimensions of force \[ [F]=[MLT^{-2}] \] Dimensions of area \[ [A]=[L^2] \] Therefore, \[ [\text{Stress}] = \frac{[MLT^{-2}]}{[L^2]} \] \[ =[ML^{-1}T^{-2}] \]
Stress is defined as \[ \text{Stress} = \frac{\text{Force}}{\text{Area}} \] Dimensions of force \[ [F]=[MLT^{-2}] \] Dimensions of area \[ [A]=[L^2] \] Therefore, \[ [\text{Stress}] = \frac{[MLT^{-2}]}{[L^2]} \] \[ =[ML^{-1}T^{-2}] \]
16
The pair of quantities having same dimensions is
Correct Option: (c)
Impulse \[ = \text{Force}\times\text{time} \] \[ =[MLT^{-2}][T] =[MLT^{-1}] \] Surface tension \[ =\frac{\text{Force}}{\text{length}} =\frac{[MLT^{-2}]}{[L]} =[MT^{-2}] \] Angular momentum \[ =\text{Moment of inertia}\times\text{angular velocity} \] \[ =[ML^2][T^{-1}] =[ML^2T^{-1}] \] Work \[ =\text{Force}\times\text{distance} \] \[ =[MLT^{-2}][L] =[ML^2T^{-2}] \] Torque \[ =\text{Force}\times\text{distance} =[ML^2T^{-2}] \] Young's modulus \[ =\frac{\text{Force/Area}} {\text{Change in length/Original length}} \] \[ =[ML^{-1}T^{-2}] \] Hence, work and torque have the same dimensions.
Impulse \[ = \text{Force}\times\text{time} \] \[ =[MLT^{-2}][T] =[MLT^{-1}] \] Surface tension \[ =\frac{\text{Force}}{\text{length}} =\frac{[MLT^{-2}]}{[L]} =[MT^{-2}] \] Angular momentum \[ =\text{Moment of inertia}\times\text{angular velocity} \] \[ =[ML^2][T^{-1}] =[ML^2T^{-1}] \] Work \[ =\text{Force}\times\text{distance} \] \[ =[MLT^{-2}][L] =[ML^2T^{-2}] \] Torque \[ =\text{Force}\times\text{distance} =[ML^2T^{-2}] \] Young's modulus \[ =\frac{\text{Force/Area}} {\text{Change in length/Original length}} \] \[ =[ML^{-1}T^{-2}] \] Hence, work and torque have the same dimensions.
17
The dimensions of \( (\mu_0\varepsilon_0)^{-1/2} \) are
Correct Option: (c)
Speed of light in vacuum, \[ c=\frac{1}{\sqrt{\mu_0\varepsilon_0}} =(\mu_0\varepsilon_0)^{-1/2} \] Therefore, \[ [(\mu_0\varepsilon_0)^{-1/2}] =[c] =[LT^{-1}] \]
Speed of light in vacuum, \[ c=\frac{1}{\sqrt{\mu_0\varepsilon_0}} =(\mu_0\varepsilon_0)^{-1/2} \] Therefore, \[ [(\mu_0\varepsilon_0)^{-1/2}] =[c] =[LT^{-1}] \]
18
The dimension of
\[
\frac{1}{2}\varepsilon_0E^2
\]
where \(\varepsilon_0\) is permittivity of free space and \(E\) is electric field, is
Correct Option: (b)
Energy density of electric field, \[ u_E=\frac12\varepsilon_0E^2 \] Energy density \[ =\frac{\text{Energy}}{\text{Volume}} \] \[ =\frac{[ML^2T^{-2}]}{[L^3]} =[ML^{-1}T^{-2}] \] Hence, \[ \left[\frac12\varepsilon_0E^2\right] =[ML^{-1}T^{-2}] \]
Energy density of electric field, \[ u_E=\frac12\varepsilon_0E^2 \] Energy density \[ =\frac{\text{Energy}}{\text{Volume}} \] \[ =\frac{[ML^2T^{-2}]}{[L^3]} =[ML^{-1}T^{-2}] \] Hence, \[ \left[\frac12\varepsilon_0E^2\right] =[ML^{-1}T^{-2}] \]
19
If the dimensions of a physical quantity are given by \(M^aL^bT^c\), then the physical quantity will be
Correct Option: (d)
Pressure, \[ P=\frac{\text{Force}}{\text{Area}} \] \[ [P] = \frac{[MLT^{-2}]}{[L^2]} \] \[ =[ML^{-1}T^{-2}] \] Comparing with \[ M^aL^bT^c \] we get \[ a=1,\quad b=-1,\quad c=-2 \]
Pressure, \[ P=\frac{\text{Force}}{\text{Area}} \] \[ [P] = \frac{[MLT^{-2}]}{[L^2]} \] \[ =[ML^{-1}T^{-2}] \] Comparing with \[ M^aL^bT^c \] we get \[ a=1,\quad b=-1,\quad c=-2 \]
20
Which two of the following five physical parameters have the same dimensions?
1. Energy density
2. Refractive index
3. Dielectric constant
4. Young's modulus
5. Magnetic field
1. Energy density
2. Refractive index
3. Dielectric constant
4. Young's modulus
5. Magnetic field
Correct Option: (a)
Energy density \[ = \frac{\text{Work done}}{\text{Volume}} \] \[ = \frac{[ML^2T^{-2}]}{[L^3]} =[ML^{-1}T^{-2}] \] Young's modulus \[ [Y] = \left[\frac{\text{Force}}{\text{Area}}\right] \times \left[\frac{l}{\Delta l}\right] \] \[ = \frac{[MLT^{-2}]}{[L^2]} \times \frac{[L]}{[L]} \] \[ =[ML^{-1}T^{-2}] \] Hence, quantities 1 and 4 have the same dimensions.
Energy density \[ = \frac{\text{Work done}}{\text{Volume}} \] \[ = \frac{[ML^2T^{-2}]}{[L^3]} =[ML^{-1}T^{-2}] \] Young's modulus \[ [Y] = \left[\frac{\text{Force}}{\text{Area}}\right] \times \left[\frac{l}{\Delta l}\right] \] \[ = \frac{[MLT^{-2}]}{[L^2]} \times \frac{[L]}{[L]} \] \[ =[ML^{-1}T^{-2}] \] Hence, quantities 1 and 4 have the same dimensions.
21
Dimensions of resistance in an electrical circuit, in terms of dimension of mass \(M\), length \(L\), time \(T\) and current \(I\), would be
Correct Option: (c)
According to Ohm's law, \[ V=RI \] or \[ R=\frac{V}{I} \] Potential difference, \[ V=\frac{W}{q} \] \[ [V] = \frac{[ML^2T^{-2}]}{[IT]} \] Therefore, \[ [R] = \frac{[ML^2T^{-2}]/[IT]}{[I]} \] \[ =[ML^2T^{-3}I^{-2}] \]
According to Ohm's law, \[ V=RI \] or \[ R=\frac{V}{I} \] Potential difference, \[ V=\frac{W}{q} \] \[ [V] = \frac{[ML^2T^{-2}]}{[IT]} \] Therefore, \[ [R] = \frac{[ML^2T^{-2}]/[IT]}{[I]} \] \[ =[ML^2T^{-3}I^{-2}] \]
22
The ratio of the dimensions of Planck's constant and that of moment of inertia is the dimensions of
Correct Option: (b)
Planck's constant \[ h=E\times t \] \[ [h] =[ML^2T^{-2}][T] =[ML^2T^{-1}] \] Moment of inertia \[ [I] =[ML^2] \] Therefore, \[ \frac{h}{I} = \frac{[ML^2T^{-1}]}{[ML^2]} \] \[ =[T^{-1}] \] which is the dimension of frequency.
Planck's constant \[ h=E\times t \] \[ [h] =[ML^2T^{-2}][T] =[ML^2T^{-1}] \] Moment of inertia \[ [I] =[ML^2] \] Therefore, \[ \frac{h}{I} = \frac{[ML^2T^{-1}]}{[ML^2]} \] \[ =[T^{-1}] \] which is the dimension of frequency.
23
The dimensions of universal gravitational constant are
Correct Option: (a)
Gravitational constant \(G\) \[ G=\frac{\text{Force}\times(\text{distance})^2} {\text{mass}\times\text{mass}} \] \[ [G] = \frac{[MLT^{-2}][L^2]} {[M][M]} \] \[ =[M^{-1}L^3T^{-2}] \]
Gravitational constant \(G\) \[ G=\frac{\text{Force}\times(\text{distance})^2} {\text{mass}\times\text{mass}} \] \[ [G] = \frac{[MLT^{-2}][L^2]} {[M][M]} \] \[ =[M^{-1}L^3T^{-2}] \]
24
The dimensions of Planck's constant equals to that of
Correct Option: (c)
Planck's constant, \[ h=\frac{\text{Energy}}{\text{Frequency}} \] \[ [h] = \frac{[ML^2T^{-2}]} {[T^{-1}]} \] \[ =[ML^2T^{-1}] \] Angular momentum, \[ L=I\omega \] \[ =[ML^2][T^{-1}] \] \[ =[ML^2T^{-1}] \] Hence Planck's constant and angular momentum have the same dimensions.
Planck's constant, \[ h=\frac{\text{Energy}}{\text{Frequency}} \] \[ [h] = \frac{[ML^2T^{-2}]} {[T^{-1}]} \] \[ =[ML^2T^{-1}] \] Angular momentum, \[ L=I\omega \] \[ =[ML^2][T^{-1}] \] \[ =[ML^2T^{-1}] \] Hence Planck's constant and angular momentum have the same dimensions.
25
Which pair do not have equal dimensions?
Correct Option: (b)
Dimensions of force \[ =[MLT^{-2}] \] Dimensions of impulse \[ =\text{Force}\times\text{time} \] \[ =[MLT^{-2}][T] =[MLT^{-1}] \] Since the dimensions are different, force and impulse do not have equal dimensions.
Dimensions of force \[ =[MLT^{-2}] \] Dimensions of impulse \[ =\text{Force}\times\text{time} \] \[ =[MLT^{-2}][T] =[MLT^{-1}] \] Since the dimensions are different, force and impulse do not have equal dimensions.
26
The dimensions of impulse are equal to that of
Correct Option: (b)
Impulse \[ =\text{Force}\times\text{Time} \] Therefore, \[ [\text{Impulse}] =[MLT^{-2}][T] \] \[ =[MLT^{-1}] \] The dimensional formula of linear momentum is also \[ [MLT^{-1}] \] Hence impulse and linear momentum have the same dimensions.
Impulse \[ =\text{Force}\times\text{Time} \] Therefore, \[ [\text{Impulse}] =[MLT^{-2}][T] \] \[ =[MLT^{-1}] \] The dimensional formula of linear momentum is also \[ [MLT^{-1}] \] Hence impulse and linear momentum have the same dimensions.
27
Which of the following dimensions will be the same as that of time?
Correct Option: (a)
Using electrical dimensions, \[ [L]=[\text{Inductance}] =[ML^2T^{-2}A^{-2}] \] \[ [R]=[\text{Resistance}] =[ML^2T^{-3}A^{-2}] \] Therefore, \[ \frac{L}{R} = \frac{[ML^2T^{-2}A^{-2}]} {[ML^2T^{-3}A^{-2}]} \] \[ =[T] \] Hence \(L/R\) has the dimensions of time.
Using electrical dimensions, \[ [L]=[\text{Inductance}] =[ML^2T^{-2}A^{-2}] \] \[ [R]=[\text{Resistance}] =[ML^2T^{-3}A^{-2}] \] Therefore, \[ \frac{L}{R} = \frac{[ML^2T^{-2}A^{-2}]} {[ML^2T^{-3}A^{-2}]} \] \[ =[T] \] Hence \(L/R\) has the dimensions of time.
28
The dimensions of \(RC\) is
Correct Option: (c)
\[ RC=(\Omega)\left(\Omega^{-1}\,s\right) \] \[ =s \] Therefore, \[ [RC]=[T] \] Hence \(RC\) has the dimensions of time.
\[ RC=(\Omega)\left(\Omega^{-1}\,s\right) \] \[ =s \] Therefore, \[ [RC]=[T] \] Hence \(RC\) has the dimensions of time.
29
Which of the following has the dimensions of pressure?
Correct Option: (b)
Pressure, \[ P=\frac{\text{Force}}{\text{Area}} \] Therefore, \[ [P] = \frac{[MLT^{-2}]} {[L^2]} \] \[ =[ML^{-1}T^{-2}] \]
Pressure, \[ P=\frac{\text{Force}}{\text{Area}} \] Therefore, \[ [P] = \frac{[MLT^{-2}]} {[L^2]} \] \[ =[ML^{-1}T^{-2}] \]
30
Of the following quantities, which one has dimensions different from the remaining three?
Correct Option: (d)
Dimensions of energy, \[ [E]=[ML^2T^{-2}] \] Dimensions of volume, \[ [V]=[L^3] \] Hence, \[ \left[\frac{E}{V}\right] = [ML^{-1}T^{-2}] \] Also, \[ \left[\frac{F}{A}\right] = [ML^{-1}T^{-2}] \] Dimensions of voltage, \[ [V]=[ML^2T^{-3}A^{-1}] \] Dimensions of charge, \[ [q]=[AT] \] \[ \left[\frac{Vq}{V}\right] =[ML^{-1}T^{-2}] \] Angular momentum, \[ [L]=[ML^2T^{-1}] \] Thus angular momentum has dimensions different from the other three quantities.
Dimensions of energy, \[ [E]=[ML^2T^{-2}] \] Dimensions of volume, \[ [V]=[L^3] \] Hence, \[ \left[\frac{E}{V}\right] = [ML^{-1}T^{-2}] \] Also, \[ \left[\frac{F}{A}\right] = [ML^{-1}T^{-2}] \] Dimensions of voltage, \[ [V]=[ML^2T^{-3}A^{-1}] \] Dimensions of charge, \[ [q]=[AT] \] \[ \left[\frac{Vq}{V}\right] =[ML^{-1}T^{-2}] \] Angular momentum, \[ [L]=[ML^2T^{-1}] \] Thus angular momentum has dimensions different from the other three quantities.
31
The dimensional formula of magnetic flux is
Correct Option: (c)
Magnetic flux, \[ \phi=BA \] Also, \[ B=\frac{F}{Il} \] Therefore, \[ \phi = \left(\frac{F}{Il}\right)A \] \[ [\phi] = \frac{[MLT^{-2}][L^2]} {[A][L]} \] \[ =[ML^2T^{-2}A^{-1}] \]
Magnetic flux, \[ \phi=BA \] Also, \[ B=\frac{F}{Il} \] Therefore, \[ \phi = \left(\frac{F}{Il}\right)A \] \[ [\phi] = \frac{[MLT^{-2}][L^2]} {[A][L]} \] \[ =[ML^2T^{-2}A^{-1}] \]
32
The dimensional formula of permeability of free space \( \mu_0 \) is
Correct Option: (a)
Permeability of free space, \[ \mu_0= \frac{2\pi\times\text{Force}\times\text{distance}} {\text{current}\times\text{current}\times\text{length}} \] Hence, \[ [\mu_0] = \frac{[MLT^{-2}][L]} {[A][A][L]} \] \[ =[MLT^{-2}A^{-2}] \]
Permeability of free space, \[ \mu_0= \frac{2\pi\times\text{Force}\times\text{distance}} {\text{current}\times\text{current}\times\text{length}} \] Hence, \[ [\mu_0] = \frac{[MLT^{-2}][L]} {[A][A][L]} \] \[ =[MLT^{-2}A^{-2}] \]
33
According to Newton, the viscous force acting between liquid layers of area \(A\) and velocity gradient \(\Delta v/\Delta z\) is given by
\[
F=-\eta A\frac{\Delta v}{\Delta z}
\]
where \(\eta\) is constant called coefficient of viscosity. The dimensional formula of \(\eta\) is
Correct Option: (d)
Given, \[ F=\eta A\frac{\Delta v}{\Delta z} \] Therefore, \[ \eta= \frac{F} {A(\Delta v/\Delta z)} \] \[ [\eta] = \frac{[MLT^{-2}]} {[L^2][T^{-1}]} \] \[ =[ML^{-1}T^{-1}] \]
Given, \[ F=\eta A\frac{\Delta v}{\Delta z} \] Therefore, \[ \eta= \frac{F} {A(\Delta v/\Delta z)} \] \[ [\eta] = \frac{[MLT^{-2}]} {[L^2][T^{-1}]} \] \[ =[ML^{-1}T^{-1}] \]
34
Dimensional formula of self inductance is
Correct Option: (c)
Induced emf, \[ |\varepsilon|=L\frac{dI}{dt} \] where \(L\) is self inductance. \[ L=\frac{|\varepsilon|}{dI/dt} \] \[ [L]=\frac{[ML^2T^{-3}A^{-1}]}{[AT^{-1}]} \] \[ =[ML^2T^{-2}A^{-2}] \]
Induced emf, \[ |\varepsilon|=L\frac{dI}{dt} \] where \(L\) is self inductance. \[ L=\frac{|\varepsilon|}{dI/dt} \] \[ [L]=\frac{[ML^2T^{-3}A^{-1}]}{[AT^{-1}]} \] \[ =[ML^2T^{-2}A^{-2}] \]
35
The dimensional formula of torque is
Correct Option: (a)
Torque, \[ \tau=\text{Force}\times\text{distance} \] \[ [\tau]=[MLT^{-2}][L] \] \[ =[ML^2T^{-2}] \]
Torque, \[ \tau=\text{Force}\times\text{distance} \] \[ [\tau]=[MLT^{-2}][L] \] \[ =[ML^2T^{-2}] \]
36
If \(C\) and \(R\) denote capacitance and resistance, the dimensional formula of \(CR\) is
Correct Option: (a)
Capacitance, \[ C=\frac{\text{Charge}}{\text{Potential difference}} \] \[ [C]=\frac{[AT]}{[ML^2T^{-3}A^{-1}]} \] \[ =[M^{-1}L^{-2}T^4A^2] \] Resistance, \[ R=\frac{\text{Potential difference}}{\text{Current}} \] \[ [R]=\frac{[ML^2T^{-3}A^{-1}]}{[A]} \] \[ =[ML^2T^{-3}A^{-2}] \] Therefore, \[ [CR] =[M^{-1}L^{-2}T^4A^2] [ML^2T^{-3}A^{-2}] =[T] \]
Capacitance, \[ C=\frac{\text{Charge}}{\text{Potential difference}} \] \[ [C]=\frac{[AT]}{[ML^2T^{-3}A^{-1}]} \] \[ =[M^{-1}L^{-2}T^4A^2] \] Resistance, \[ R=\frac{\text{Potential difference}}{\text{Current}} \] \[ [R]=\frac{[ML^2T^{-3}A^{-1}]}{[A]} \] \[ =[ML^2T^{-3}A^{-2}] \] Therefore, \[ [CR] =[M^{-1}L^{-2}T^4A^2] [ML^2T^{-3}A^{-2}] =[T] \]
37
The dimensional formula of angular momentum is
Correct Option: (d)
Angular momentum, \[ L=I\omega \] where \(I\) is moment of inertia. \[ [L]=[ML^2][T^{-1}] \] \[ =[ML^2T^{-1}] \]
Angular momentum, \[ L=I\omega \] where \(I\) is moment of inertia. \[ [L]=[ML^2][T^{-1}] \] \[ =[ML^2T^{-1}] \]
38
A physical quantity of the dimensions of length that can be formed out of \(c\), \(G\) and
\[
\frac{e^2}{4\pi\varepsilon_0}
\]
is \([c\text{ is velocity of light, }G\text{ is the universal constant of gravitation and }e\text{ is charge}]\)
Correct Option: (d)
\[ \left[\frac{e^2}{4\pi\varepsilon_0}\right] =[F\times d^2] \] \[ =[ML^3T^{-2}] \] Also, \[ [G]=[M^{-1}L^3T^{-2}] \] and \[ [c]=[LT^{-1}] \] Let \[ l\propto \left(\frac{e^2}{4\pi\varepsilon_0}\right)^p G^q c^r \] Comparing dimensions, \[ [L] =[ML^3T^{-2}]^p [M^{-1}L^3T^{-2}]^q [LT^{-1}]^r \] Solving, \[ p=\frac12,\quad q=\frac12,\quad r=-2 \] Therefore, \[ l\propto \frac{1}{c^2} \left( G\frac{e^2}{4\pi\varepsilon_0} \right)^{1/2} \]
\[ \left[\frac{e^2}{4\pi\varepsilon_0}\right] =[F\times d^2] \] \[ =[ML^3T^{-2}] \] Also, \[ [G]=[M^{-1}L^3T^{-2}] \] and \[ [c]=[LT^{-1}] \] Let \[ l\propto \left(\frac{e^2}{4\pi\varepsilon_0}\right)^p G^q c^r \] Comparing dimensions, \[ [L] =[ML^3T^{-2}]^p [M^{-1}L^3T^{-2}]^q [LT^{-1}]^r \] Solving, \[ p=\frac12,\quad q=\frac12,\quad r=-2 \] Therefore, \[ l\propto \frac{1}{c^2} \left( G\frac{e^2}{4\pi\varepsilon_0} \right)^{1/2} \]
39
Planck's constant (\(h\)), speed of light in vacuum (\(c\)) and Newton's gravitational constant (\(G\)) are three fundamental constants. Which of the following combinations of these has the dimension of length?
Correct Option: (a)
Let \[ l= \frac{\sqrt{hG}}{c^{3/2}} \] Using \[ [h]=[ML^2T^{-1}] \] \[ [G]=[M^{-1}L^3T^{-2}] \] \[ [c]=[LT^{-1}] \] Therefore, \[ [l] = \frac{ \sqrt{[ML^2T^{-1}] [M^{-1}L^3T^{-2}]} } {[LT^{-1}]^{3/2}} \] \[ =[L] \] Hence option (a) has dimensions of length.
Let \[ l= \frac{\sqrt{hG}}{c^{3/2}} \] Using \[ [h]=[ML^2T^{-1}] \] \[ [G]=[M^{-1}L^3T^{-2}] \] \[ [c]=[LT^{-1}] \] Therefore, \[ [l] = \frac{ \sqrt{[ML^2T^{-1}] [M^{-1}L^3T^{-2}]} } {[LT^{-1}]^{3/2}} \] \[ =[L] \] Hence option (a) has dimensions of length.
40
If dimensions of critical velocity \(v_c\) of a liquid flowing through a tube are expressed as
\[
[\eta^x \rho^y r^z]
\]
where \(\eta\), \(\rho\) and \(r\) are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of \(x, y\) and \(z\) are given by
Correct Option: (d)
Let \[ v_c\propto \eta^x\rho^y r^z \] Dimensions are \[ [v_c]=[LT^{-1}] \] \[ [\eta]=[ML^{-1}T^{-1}] \] \[ [\rho]=[ML^{-3}] \] \[ [r]=[L] \] Therefore, \[ [LT^{-1}] = [ML^{-1}T^{-1}]^x [ML^{-3}]^y [L]^z \] Comparing powers, \[ x=-1,\quad y=-1,\quad z=1 \]
Let \[ v_c\propto \eta^x\rho^y r^z \] Dimensions are \[ [v_c]=[LT^{-1}] \] \[ [\eta]=[ML^{-1}T^{-1}] \] \[ [\rho]=[ML^{-3}] \] \[ [r]=[L] \] Therefore, \[ [LT^{-1}] = [ML^{-1}T^{-1}]^x [ML^{-3}]^y [L]^z \] Comparing powers, \[ x=-1,\quad y=-1,\quad z=1 \]
41
If force \(F\), velocity \(V\) and time \(T\) are taken as fundamental units, then the dimensions of mass are
Correct Option: (d)
Let \[ m\propto F^aV^bT^c \] Then, \[ [M] =[MLT^{-2}]^a [LT^{-1}]^b [T]^c \] Equating powers, \[ a=1,\quad a+b=0,\quad -2a-b+c=0 \] Hence, \[ a=1,\quad b=-1,\quad c=1 \] Therefore, \[ [m]=[FV^{-1}T] \]
Let \[ m\propto F^aV^bT^c \] Then, \[ [M] =[MLT^{-2}]^a [LT^{-1}]^b [T]^c \] Equating powers, \[ a=1,\quad a+b=0,\quad -2a-b+c=0 \] Hence, \[ a=1,\quad b=-1,\quad c=1 \] Therefore, \[ [m]=[FV^{-1}T] \]
42
The density of a material in CGS system of units is \(4\,g\,cm^{-3}\). In a system of units in which unit of length is \(10\,cm\) and unit of mass is \(100\,g\), the value of density of material will be
Correct Option: (c)
Given, \[ \rho_1=4\,g\,cm^{-3} \] New unit of mass \[ =100g \] New unit of length \[ =10cm \] Hence new unit of density, \[ \frac{100g}{(10cm)^3} =0.1\,g\,cm^{-3} \] Therefore, \[ n_2=\frac{4}{0.1}=40 \]
Given, \[ \rho_1=4\,g\,cm^{-3} \] New unit of mass \[ =100g \] New unit of length \[ =10cm \] Hence new unit of density, \[ \frac{100g}{(10cm)^3} =0.1\,g\,cm^{-3} \] Therefore, \[ n_2=\frac{4}{0.1}=40 \]
43
The velocity \(v\) of a particle at time \(t\) is given by
\[
v=at+\frac{b}{t+c}
\]
where \(a,b\) and \(c\) are constants. The dimensions of \(a,b\) and \(c\) are
Correct Option: (b)
Given, \[ v=at+\frac{b}{t+c} \] Since \(t\) and \(c\) are added, \[ [c]=[T] \] Also, \[ [at]=[LT^{-1}] \] Therefore, \[ [a]=\frac{[LT^{-1}]}{[T]} =[LT^{-2}] \] Further, \[ \frac{[b]}{[T]}=[LT^{-1}] \] Hence, \[ [b]=[L] \]
Given, \[ v=at+\frac{b}{t+c} \] Since \(t\) and \(c\) are added, \[ [c]=[T] \] Also, \[ [at]=[LT^{-1}] \] Therefore, \[ [a]=\frac{[LT^{-1}]}{[T]} =[LT^{-2}] \] Further, \[ \frac{[b]}{[T]}=[LT^{-1}] \] Hence, \[ [b]=[L] \]
44
An equation is given here
\[
\left(P+\frac{a}{V^2}\right)=b\frac{\theta}{V}
\]
where \(P\) = Pressure, \(V\) = Volume and \(\theta\) = Absolute temperature. If \(a\) and \(b\) are constants, the dimensions of \(a\) will be
Correct Option: (c)
Given, \[ \left(P+\frac{a}{V^2}\right) = b\frac{\theta}{V} \] Since \[ \frac{a}{V^2} \] is added to pressure \(P\), \[ \left[\frac{a}{V^2}\right] =[P] \] Therefore, \[ [a] =[P][V^2] \] \[ =[ML^{-1}T^{-2}] [L^3]^2 \] \[ =[ML^5T^{-2}] \]
Given, \[ \left(P+\frac{a}{V^2}\right) = b\frac{\theta}{V} \] Since \[ \frac{a}{V^2} \] is added to pressure \(P\), \[ \left[\frac{a}{V^2}\right] =[P] \] Therefore, \[ [a] =[P][V^2] \] \[ =[ML^{-1}T^{-2}] [L^3]^2 \] \[ =[ML^5T^{-2}] \]
45
Which of the following is a dimensional constant?
Correct Option: (b)
Relative density, refractive index and Poisson's ratio are ratios and hence dimensionless constants. Gravitational constant \(G\) has dimensions. Therefore, the dimensional constant is gravitational constant.
Relative density, refractive index and Poisson's ratio are ratios and hence dimensionless constants. Gravitational constant \(G\) has dimensions. Therefore, the dimensional constant is gravitational constant.
46
Turpentine oil is flowing through a tube of length \(l\) and radius \(r\). The pressure difference between the two ends of the tube is \(P\). The viscosity of oil is given by
\[
\eta=\frac{P(r^2-x^2)}{4vl}
\]
where \(v\) is the velocity of oil at a distance \(x\) from the axis of the tube. The dimensions of \(\eta\) are
Correct Option: (d)
Dimensions of pressure, \[ [P]=[ML^{-1}T^{-2}] \] Dimensions of \(r\), \[ [r]=[L] \] Dimensions of velocity, \[ [v]=[LT^{-1}] \] Dimensions of length, \[ [l]=[L] \] Given, \[ \eta=\frac{P(r^2-x^2)}{4vl} \] Therefore, \[ [\eta] = \frac{[ML^{-1}T^{-2}][L^2]} {[LT^{-1}][L]} \] \[ =[ML^{-1}T^{-1}] \]
Dimensions of pressure, \[ [P]=[ML^{-1}T^{-2}] \] Dimensions of \(r\), \[ [r]=[L] \] Dimensions of velocity, \[ [v]=[LT^{-1}] \] Dimensions of length, \[ [l]=[L] \] Given, \[ \eta=\frac{P(r^2-x^2)}{4vl} \] Therefore, \[ [\eta] = \frac{[ML^{-1}T^{-2}][L^2]} {[LT^{-1}][L]} \] \[ =[ML^{-1}T^{-1}] \]
47
The time dependence of a physical quantity \(p\) is given by
\[
p=p_0\exp(-at^2)
\]
where \(a\) is a constant and \(t\) is the time. The constant \(a\)
Correct Option: (b)
Given, \[ p=p_0e^{-\alpha t^2} \] Since the exponent of an exponential function must be dimensionless, \[ \alpha t^2 \] is dimensionless. Therefore, \[ [\alpha]=\frac{1}{[T^2]} =[T^{-2}] \]
Given, \[ p=p_0e^{-\alpha t^2} \] Since the exponent of an exponential function must be dimensionless, \[ \alpha t^2 \] is dimensionless. Therefore, \[ [\alpha]=\frac{1}{[T^2]} =[T^{-2}] \]
48
\(P\) represents radiation pressure, \(c\) represents speed of light and \(S\) represents radiation energy striking per unit area per sec. The non-zero integers \(x,y,z\) such that
\[
P^xS^yc^z
\]
is dimensionless are
Correct Option: (c)
Let \[ k=P^xS^yc^z \] where \(k\) is dimensionless. Dimensions of pressure, \[ [P]=\frac{\text{Force}}{\text{Area}} =[ML^{-1}T^{-2}] \] Dimensions of radiation energy per unit area per second, \[ [S] = \frac{\text{Energy}} {\text{Area}\times\text{time}} =[MT^{-3}] \] Dimensions of speed of light, \[ [c]=[LT^{-1}] \] Therefore, \[ [M^0L^0T^0] =[ML^{-1}T^{-2}]^x [MT^{-3}]^y [LT^{-1}]^z \] Comparing powers, \[ x+y=0 \] \[ -x+z=0 \] \[ -2x-3y-z=0 \] Solving, \[ x=1,\quad y=-1,\quad z=1 \]
Let \[ k=P^xS^yc^z \] where \(k\) is dimensionless. Dimensions of pressure, \[ [P]=\frac{\text{Force}}{\text{Area}} =[ML^{-1}T^{-2}] \] Dimensions of radiation energy per unit area per second, \[ [S] = \frac{\text{Energy}} {\text{Area}\times\text{time}} =[MT^{-3}] \] Dimensions of speed of light, \[ [c]=[LT^{-1}] \] Therefore, \[ [M^0L^0T^0] =[ML^{-1}T^{-2}]^x [MT^{-3}]^y [LT^{-1}]^z \] Comparing powers, \[ x+y=0 \] \[ -x+z=0 \] \[ -2x-3y-z=0 \] Solving, \[ x=1,\quad y=-1,\quad z=1 \]
49
The frequency of vibration \(f\) of a mass \(m\) suspended from a spring of spring constant \(k\) is given by a relation
\[
f=am^xk^y
\]
where \(a\) is a dimensionless constant. The values of \(x\) and \(y\) are
Correct Option: (d)
Given, \[ f=am^xk^y \] where \(a\) is dimensionless. Dimensions of frequency, \[ [f]=[T^{-1}] \] Dimensions of mass, \[ [m]=[M] \] Dimensions of spring constant, \[ [k]=[MT^{-2}] \] Therefore, \[ [T^{-1}] =[M]^x[MT^{-2}]^y \] Comparing powers, \[ x+y=0 \] \[ -2y=-1 \] Hence, \[ y=\frac12, \qquad x=-\frac12 \]
Given, \[ f=am^xk^y \] where \(a\) is dimensionless. Dimensions of frequency, \[ [f]=[T^{-1}] \] Dimensions of mass, \[ [m]=[M] \] Dimensions of spring constant, \[ [k]=[MT^{-2}] \] Therefore, \[ [T^{-1}] =[M]^x[MT^{-2}]^y \] Comparing powers, \[ x+y=0 \] \[ -2y=-1 \] Hence, \[ y=\frac12, \qquad x=-\frac12 \]
50
If
\[
x=at+bt^2
\]
where \(x\) is the distance travelled by the body in kilometers while \(t\) is the time in seconds, then the units of \(b\) is
Correct Option: (c)
Given, \[ x=at+bt^2 \] where \(x\) is measured in km and \(t\) in seconds. Therefore, \[ b=\frac{x}{t^2} \] Hence the unit of \(b\) is \[ \frac{\text{km}}{\text{s}^2} \]
Given, \[ x=at+bt^2 \] where \(x\) is measured in km and \(t\) in seconds. Therefore, \[ b=\frac{x}{t^2} \] Hence the unit of \(b\) is \[ \frac{\text{km}}{\text{s}^2} \]
51
A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that, 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that, 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is
Correct Option: (d)
Circular scale reading, \[ \text{CSR}=52 \] Least count, \[ LC=\frac{p}{n} \] Here, \[ p=1\text{ mm},\qquad n=100 \] Therefore, \[ LC=\frac{1}{100}\text{ mm} =0.01\text{ mm} =0.001\text{ cm} \] Diameter of wire, \[ D=\text{MSR}+(\text{CSR}\times LC) \] \[ D=0+(52\times0.001) \] \[ D=0.052\text{ cm} \]
Circular scale reading, \[ \text{CSR}=52 \] Least count, \[ LC=\frac{p}{n} \] Here, \[ p=1\text{ mm},\qquad n=100 \] Therefore, \[ LC=\frac{1}{100}\text{ mm} =0.01\text{ mm} =0.001\text{ cm} \] Diameter of wire, \[ D=\text{MSR}+(\text{CSR}\times LC) \] \[ D=0+(52\times0.001) \] \[ D=0.052\text{ cm} \]
52
If force \([F]\), acceleration \([a]\) and time \([T]\) are chosen as the fundamental physical quantities. Find the dimensions of energy.
Correct Option: (b)
Fundamental quantities are force \([F]\), acceleration \([a]\) and time \([T]\). Let, \[ [E]=[F]^x[a]^y[T]^z \] Since, \[ [E]=[ML^2T^{-2}] \] \[ [F]=[MLT^{-2}] \] \[ [a]=[LT^{-2}] \] Therefore, \[ [ML^2T^{-2}] =[MLT^{-2}]^x [LT^{-2}]^y [T]^z \] Comparing powers, \[ x=1,\qquad x+y=2,\qquad -2x-2y+z=-2 \] Solving, \[ x=1,\quad y=1,\quad z=2 \] Hence, \[ [E]=[F][a][T^2] \]
Fundamental quantities are force \([F]\), acceleration \([a]\) and time \([T]\). Let, \[ [E]=[F]^x[a]^y[T]^z \] Since, \[ [E]=[ML^2T^{-2}] \] \[ [F]=[MLT^{-2}] \] \[ [a]=[LT^{-2}] \] Therefore, \[ [ML^2T^{-2}] =[MLT^{-2}]^x [LT^{-2}]^y [T]^z \] Comparing powers, \[ x=1,\qquad x+y=2,\qquad -2x-2y+z=-2 \] Solving, \[ x=1,\quad y=1,\quad z=2 \] Hence, \[ [E]=[F][a][T^2] \]
53
If \(E\) and \(G\) respectively denote energy and gravitational constant, then
\[
\frac{E}{G}
\]
has the dimensions of
Correct Option: (a)
Dimensions of energy, \[ [E]=[ML^2T^{-2}] \] From Newton's law of gravitation, \[ F=\frac{GM_1M_2}{r^2} \] \[ G=\frac{Fr^2}{M_1M_2} \] Therefore, \[ [G] = \frac{[MLT^{-2}][L^2]} {[M][M]} \] \[ =[M^{-1}L^3T^{-2}] \] Hence, \[ \left[\frac{E}{G}\right] = \frac{[ML^2T^{-2}]} {[M^{-1}L^3T^{-2}]} \] \[ =[M^2L^{-1}] \]
Dimensions of energy, \[ [E]=[ML^2T^{-2}] \] From Newton's law of gravitation, \[ F=\frac{GM_1M_2}{r^2} \] \[ G=\frac{Fr^2}{M_1M_2} \] Therefore, \[ [G] = \frac{[MLT^{-2}][L^2]} {[M][M]} \] \[ =[M^{-1}L^3T^{-2}] \] Hence, \[ \left[\frac{E}{G}\right] = \frac{[ML^2T^{-2}]} {[M^{-1}L^3T^{-2}]} \] \[ =[M^2L^{-1}] \]
54
Plane angle and solid angle have :
Correct Option: (a)
Plane angle and solid angle are dimensionless quantities though they possess units. Dimensions of the given quantities are: \[ \text{Magnetic flux}=[ML^2T^{-2}A^{-1}] \] \[ \text{Self inductance}=[ML^2T^{-2}A^{-2}] \] \[ \text{Magnetic permeability}=[MLT^{-2}A^{-2}] \] \[ \text{Electric permittivity}=[M^{-1}L^{-3}T^4A^2] \] Therefore, plane angle is dimensionless.
Plane angle and solid angle are dimensionless quantities though they possess units. Dimensions of the given quantities are: \[ \text{Magnetic flux}=[ML^2T^{-2}A^{-1}] \] \[ \text{Self inductance}=[ML^2T^{-2}A^{-2}] \] \[ \text{Magnetic permeability}=[MLT^{-2}A^{-2}] \] \[ \text{Electric permittivity}=[M^{-1}L^{-3}T^4A^2] \] Therefore, plane angle is dimensionless.
55
The dimensions \([MLT^{-2}A^{-2}]\) belong to the:
Correct Option: (c)
Magnetic permeability, \[ \mu=\frac{B}{H} \] Since, \[ B=\frac{N}{Am} \quad \text{and} \quad H=\frac{A}{m} \] Therefore, \[ \mu=\frac{N/A m}{A/m} \] \[ =\frac{N}{A^2} \] \[ =\frac{[MLT^{-2}]}{[A^2]} \] \[ =[MLT^{-2}A^{-2}] \]
Magnetic permeability, \[ \mu=\frac{B}{H} \] Since, \[ B=\frac{N}{Am} \quad \text{and} \quad H=\frac{A}{m} \] Therefore, \[ \mu=\frac{N/A m}{A/m} \] \[ =\frac{N}{A^2} \] \[ =\frac{[MLT^{-2}]}{[A^2]} \] \[ =[MLT^{-2}A^{-2}] \]
56
The area of a rectangular field (in \(m^2\)) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digits is :
Correct Option: (d)
Area \[ =55.3\times25 \] \[ =1382.5\,m^2 \] The least number of significant figures is 2. Therefore the answer must contain 2 significant figures. \[ 1382.5 \approx 1.4\times10^3 \] \[ =14\times10^2\,m^2 \]
Area \[ =55.3\times25 \] \[ =1382.5\,m^2 \] The least number of significant figures is 2. Therefore the answer must contain 2 significant figures. \[ 1382.5 \approx 1.4\times10^3 \] \[ =14\times10^2\,m^2 \]
57
The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are
Correct Option: (c)
Random errors arise due to unpredictable variations during measurements. Since the cause of error is unpredictable, the error is classified as a random error.
Random errors arise due to unpredictable variations during measurements. Since the cause of error is unpredictable, the error is classified as a random error.
58
A metal wire has mass \((0.4\pm0.002)\,g\), radius \((0.3\pm0.001)\,mm\) and length \((5\pm0.02)\,cm\). The maximum possible percentage error in the measurement of density will nearly be
Correct Option: (b)
Density, \[ d=\frac{m}{\pi r^2l} \] Therefore, \[ \left(\frac{\Delta d}{d}\right)\times100 = \left( \frac{\Delta m}{m} +2\frac{\Delta r}{r} +\frac{\Delta l}{l} \right)\times100 \] Substituting values, \[ = \left( \frac{0.002}{0.4} +2\frac{0.001}{0.3} +\frac{0.02}{5} \right)\times100 \] \[ =1.6\% \]
Density, \[ d=\frac{m}{\pi r^2l} \] Therefore, \[ \left(\frac{\Delta d}{d}\right)\times100 = \left( \frac{\Delta m}{m} +2\frac{\Delta r}{r} +\frac{\Delta l}{l} \right)\times100 \] Substituting values, \[ = \left( \frac{0.002}{0.4} +2\frac{0.001}{0.3} +\frac{0.02}{5} \right)\times100 \] \[ =1.6\% \]
59
In a vernier callipers, \((N+1)\) divisions of vernier scale coincide with \(N\) divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:
Correct Option: (d)
Given, \[ (N+1)VSD=NMSD \] Therefore, \[ 1VSD=\frac{N}{N+1}MSD \] Vernier constant, \[ LC=1MSD-1VSD \] \[ =\left(1-\frac{N}{N+1}\right)MSD \] \[ =\frac{1}{N+1}MSD \] Since, \[ 1MSD=0.1\,cm \] \[ LC=\frac{1}{100(N+1)}cm \]
Given, \[ (N+1)VSD=NMSD \] Therefore, \[ 1VSD=\frac{N}{N+1}MSD \] Vernier constant, \[ LC=1MSD-1VSD \] \[ =\left(1-\frac{N}{N+1}\right)MSD \] \[ =\frac{1}{N+1}MSD \] Since, \[ 1MSD=0.1\,cm \] \[ LC=\frac{1}{100(N+1)}cm \]
60
The quantities which have the same dimensions as those of solid angle are:
Correct Option: (c)
Solid angle, strain and plane angle are ratios. Therefore all are dimensionless quantities.
Solid angle, strain and plane angle are ratios. Therefore all are dimensionless quantities.
61
A force defined by
\[
F=\alpha t^2+\beta t
\]
acts on a particle at a given time \(t\). The factor which is dimensionless, if \(\alpha\) and \(\beta\) are constants, is:
Correct Option: (d)
Given, \[ F=\alpha t^2+\beta t \] Dimensions of force, \[ [F]=[MLT^{-2}] \] Hence, \[ [\alpha t^2]=[MLT^{-2}] \] \[ [\alpha]=[MLT^{-4}] \] Also, \[ [\beta t]=[MLT^{-2}] \] \[ [\beta]=[MLT^{-3}] \] Therefore, \[ \left[\frac{\alpha t}{\beta}\right] = \frac{[MLT^{-4}][T]} {[MLT^{-3}]} \] \[ =[M^0L^0T^0] \] Hence it is dimensionless.
Given, \[ F=\alpha t^2+\beta t \] Dimensions of force, \[ [F]=[MLT^{-2}] \] Hence, \[ [\alpha t^2]=[MLT^{-2}] \] \[ [\alpha]=[MLT^{-4}] \] Also, \[ [\beta t]=[MLT^{-2}] \] \[ [\beta]=[MLT^{-3}] \] Therefore, \[ \left[\frac{\alpha t}{\beta}\right] = \frac{[MLT^{-4}][T]} {[MLT^{-3}]} \] \[ =[M^0L^0T^0] \] Hence it is dimensionless.
62
A balloon is made of a material of surface tension \(S\) and its inflation outlet has small area \(A\). It is filled with a gas of density \(\rho\) and takes a spherical shape of radius \(R\). When the gas is allowed to flow freely out of it, its radius \(r\) changes from \(R\) to 0 in time \(T\). If
\[
T\propto S^\alpha A^\beta \rho^\gamma R^\delta
\]
then
Correct Option: (c)
Surface tension, \[ S=\frac{F}{l} \] \[ [S]=[MT^{-2}] \] Let, \[ T\propto S^\alpha A^\beta \rho^\gamma R^\delta \] Then, \[ [T] =[MT^{-2}]^\alpha [L^2]^\beta [ML^{-3}]^\gamma [L]^\delta \] Comparing dimensions, \[ \alpha+\gamma=0 \] \[ 2\beta-3\gamma+\delta=0 \] \[ -2\alpha=1 \] Solving, \[ \alpha=-\frac12, \qquad \gamma=\frac12 \] One suitable solution is \[ \beta=-1, \qquad \delta=\frac72 \]
Surface tension, \[ S=\frac{F}{l} \] \[ [S]=[MT^{-2}] \] Let, \[ T\propto S^\alpha A^\beta \rho^\gamma R^\delta \] Then, \[ [T] =[MT^{-2}]^\alpha [L^2]^\beta [ML^{-3}]^\gamma [L]^\delta \] Comparing dimensions, \[ \alpha+\gamma=0 \] \[ 2\beta-3\gamma+\delta=0 \] \[ -2\alpha=1 \] Solving, \[ \alpha=-\frac12, \qquad \gamma=\frac12 \] One suitable solution is \[ \beta=-1, \qquad \delta=\frac72 \]
63
Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 VSD are equal to its 9 MSD. The least division in the M.S. is 0.1 cm and the zero of V.S. is at \(x=0.1\) cm when the jaws are closed. If the main scale reading is \(M=5\) cm and the number of coinciding Vernier division is 8, the measured diameter after zero error correction is
Correct Option: (c)
Given, \[ 10VSD=9MSD \] Therefore, \[ 1VSD=\frac{9}{10}MSD \] Least count, \[ LC=1MSD-1VSD \] \[ =1MSD-\frac{9}{10}MSD \] \[ =\frac{1}{10}MSD \] Since, \[ 1MSD=0.1\,cm \] \[ LC=0.01\,cm \] Reading, \[ =5+8(LC)-0.1 \] \[ =5+8(0.01)-0.1 \] \[ =4.98\,cm \]
Given, \[ 10VSD=9MSD \] Therefore, \[ 1VSD=\frac{9}{10}MSD \] Least count, \[ LC=1MSD-1VSD \] \[ =1MSD-\frac{9}{10}MSD \] \[ =\frac{1}{10}MSD \] Since, \[ 1MSD=0.1\,cm \] \[ LC=0.01\,cm \] Reading, \[ =5+8(LC)-0.1 \] \[ =5+8(0.01)-0.1 \] \[ =4.98\,cm \]
64
A physical quantity \(P\) is related to four observations \(a,b,c\) and \(d\) as follows:
\[
P=\frac{a^3b^2}{c\sqrt{d}}
\]
The percentage errors of measurement in \(a,b,c\) and \(d\) are 1%, 3%, 2%, and 4%, respectively. The percentage error in the quantity \(P\) is
Correct Option: (c)
Given, \[ P=\frac{a^3b^2}{c\sqrt d} \] Percentage error, \[ \frac{\Delta P}{P}\times100 = \left( \frac{3\Delta a}{a} +\frac{2\Delta b}{b} +\frac{\Delta c}{c} +\frac{1}{2}\frac{\Delta d}{d} \right)\times100 \] Substituting the given percentage errors, \[ =3(1\%) +2(3\%) +2\% +\frac12(4\%) \] \[ =3+6+2+2 \] \[ =13\% \]
Given, \[ P=\frac{a^3b^2}{c\sqrt d} \] Percentage error, \[ \frac{\Delta P}{P}\times100 = \left( \frac{3\Delta a}{a} +\frac{2\Delta b}{b} +\frac{\Delta c}{c} +\frac{1}{2}\frac{\Delta d}{d} \right)\times100 \] Substituting the given percentage errors, \[ =3(1\%) +2(3\%) +2\% +\frac12(4\%) \] \[ =3+6+2+2 \] \[ =13\% \]
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