Class 11 Physics - Elasticity (Important Q&A)
Elasticity Marks · 2M 3M
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📘 2 Mark Q&A
1+ problemsSolution:
Strain is more fundamental than stress.
Because Strain represents the actual deformation produced in a body, which is the direct physical effect of an applied force.
Stress is the cause (internal restoring force per unit area), whereas strain is the effect (change in dimension).
Also, strain is a dimensionless quantity and depends only on the change in configuration of the body.
Therefore, strain is considered more fundamental than stress.
📗 3 Mark Q&A
30+ problemsSolution:
For a uniform spring, spring constant \(k\) is inversely proportional to its length.
\( k \propto \dfrac{1}{L} \)
Let the original length be \(L\) and spring constant be \(k\).
The spring is cut in the ratio \(1 : 2\).
So, lengths of the two parts are \( \dfrac{L}{3} \) and \( \dfrac{2L}{3} \)
Since \( k \propto \dfrac{1}{L} \),
Spring constant of first part,
\( k_1 = k \times \dfrac{L}{L/3} \)
\( \therefore k_1 = 3k \).
Spring constant of second part,
\( k_2 = k \times \dfrac{L}{2L/3} \)
\( \therefore k_2 = \dfrac{3k}{2} \)
Answer: Therefore, the spring constants are \( 3k \) and \( \dfrac{3k}{2} \).
Solution:
For a uniform spring, spring constant \(k\) is inversely proportional to its length.
\( k \propto \dfrac{1}{L} \).
Let the original length be \(L\) and spring constant be \(k\).
The spring is cut into two equal parts, so length of each part = \( \dfrac{L}{2} \).
Since \( k \propto \dfrac{1}{L} \),
Spring constant of each part,
\( k' = k \times \dfrac{L}{L/2} \).
\( \therefore k' = 2k \).
Answer: Therefore, the spring constant of each half is \( 2k \).
Solution:
Since one end of a spring is connected to one end of the other, the springs are in series.
Let a force \(F\) be applied to the system.
Let extensions of the two springs be \(x_1\) and \(x_2\).
Then, \( F = k_1 x_1 \) and \( F = k_2 x_2 \)
Total extension of the system,
\( x = x_1 + x_2 \)
\( \Rightarrow x = \dfrac{F}{k_1} + \dfrac{F}{k_2} \)
\( \therefore x = F \left( \dfrac{1}{k_1} + \dfrac{1}{k_2} \right) \)
If \(k\) is the equivalent spring constant, then \( F = kx \)
\( x = \dfrac{F}{k} \)
On comparing the above value of \(x \), we get
\( \dfrac{F}{k} = F \left( \dfrac{1}{k_1} + \dfrac{1}{k_2} \right) \)
\( \Rightarrow \dfrac{1}{k} = \dfrac{1}{k_1} + \dfrac{1}{k_2} \)
\( \therefore k = \dfrac{k_1 k_2}{k_1 + k_2} \)
Answer: Therefore, the equivalent spring constant of the system is \( \dfrac{k_1 k_2}{k_1 + k_2} \).
Solution:
Consider a rod of original length \(L\) and cross-sectional area \(A\).
Let a force \(F\) produce an extension \(x\).
Work done in stretching the rod = elastic potential energy stored.
Since force increases gradually from \(0\) to \(F\), average force = \( \dfrac{F}{2} \).
So, work done \( W = \dfrac{F}{2} \times x \)
\( \therefore W = \dfrac{1}{2} Fx \)
Volume of the rod \( V = AL \)
Now, the elastic potential energy per unit volume is given by -
\( U = \dfrac{W}{V} = \dfrac{1}{2} \dfrac{Fx}{AL} \)
Now, stress \( = \dfrac{F}{A} \) and strain \( = \dfrac{x}{L} \)
On substitution stress and strain, we get
\( U = \dfrac{1}{2} \times \text{stress} \times \text{strain} \)
Hence proved.
Solution:
Let the original length of the wire be \(L\) and cross-sectional area be \(A\).
Let extension produced be \(x\).
Stress applied \( \sigma = \dfrac{F}{A} \)
Strain produced \( = \dfrac{x}{L} \)
From Young’s modulus,
\( Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{\sigma}{x/L} \)
So, \( \dfrac{x}{L} = \dfrac{\sigma}{Y} \)
Work done in stretching the wire = elastic potential energy stored
Since force increases gradually from \(0\) to \(F\), average force = \( \dfrac{F}{2} \)
\( \therefore W = \dfrac{1}{2} F x \)
Volume of the wire \( V = AL \)
Potential energy density \( U = \dfrac{W}{V} = \dfrac{1}{2} \dfrac{F x}{AL} \)
Substituting \( F = \sigma A \), we get
\( U = \dfrac{1}{2} \dfrac{\sigma A \, x}{AL} \)
\( \Rightarrow U = \dfrac{1}{2} \sigma \dfrac{x}{L} \)
But \( \dfrac{x}{L} = \dfrac{\sigma}{Y} \),
So, \( U = \dfrac{1}{2} \sigma \left( \dfrac{\sigma}{Y} \right) \)
\( \therefore U = \dfrac{\sigma^2}{2Y} \)
Hence proved.
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