Question 1:
easyA wire of length $L$ area of cross section $A$ is hanging from a fixed support. The length of the wire changes to $L_1$ when mass $M$ is suspended from its free end. The expression for Young’s modulus is:
(2020)
Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{Mg/A}{\Delta L/L}$. Here, the change in length $\Delta L = L_1 - L$. Substituting this gives $Y = \frac{MgL}{A(L_1 - L)}$.
Question 2:
moderateTwo wires are made of the same material and have the same volume. The first wire has cross-sectional area $A$ and the second wire has cross-sectional area $3A$. If the length of the first wire is increased by $\Delta l$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount?
(2018)
Volume $V = A_1 L_1 = A_2 L_2 \Rightarrow A L_1 = 3A L_2 \Rightarrow L_2 = L_1/3$. Force $F = \frac{Y A \Delta l}{L_1}$. For the second wire, $F' = \frac{Y (3A) \Delta l}{L_1/3} = 9 \left( \frac{Y A \Delta l}{L_1} \right) = 9F$.
Question 3:
easyThe Young’s modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of:
(2015 Re)
We know $\Delta L = \frac{FL}{AY}$. Since $L$, $A$, and $\Delta L$ are the same for both wires, $F \propto Y$. Therefore, $$\frac{F_s}{F_b} = \frac{Y_s}{Y_b} = \frac{2}{1} = 2:1$$.
Question 4:
easyCopper of fixed volume $V$ is drawn into wire of length $l$. When this wire is subjected to a constant force $F$, the extension produced in the wire is $\Delta l$. Which of the following graphs is a straight line?
(2014)
Extension $\Delta l = \frac{Fl}{AY}$. Since volume $V = Al$, we have $A = V/l$. Substituting this, $$\Delta l = \frac{Fl}{(V/l)Y} = \frac{Fl^2}{VY}$$. Thus, $\Delta l \propto l^2$, meaning the graph of $\Delta l$ versus $l^2$ is a straight line.
Question 5:
easyThe following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
(2013)
Extension $\Delta L = \frac{FL}{AY} = \frac{4FL}{\pi d^2 Y}$. For a given force and material, $$\Delta L \propto \frac{L}{d^2}$$. Calculating $L/d^2$ for the options reveals that option (b) gives the maximum value ($50 / 0.5^2 = 200$).
Question 6:
easyGiven below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R)
Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring
Reason (R): A coil spring of copper has more tensile strength than a steel spring of same dimensions.
In the light of the above statements, choose the most appropriate answer from the options given below :
(2022)
When a spring is stretched, the wire itself undergoes torsion, which is governed by the shear modulus of the material, making the assertion true. Steel has a higher tensile strength and elasticity than copper, so the reason is false.
Question 7:
moderateThe bulk modulus of a spherical objects is ‘$B$’. If it is subjected to uniform pressure ‘$P$’, the fractional decrease in radius is:
(2017-Delhi)
Bulk modulus $B = \frac{P}{\Delta V/V} \Rightarrow \frac{\Delta V}{V} = \frac{P}{B}$. For a sphere, $V = \frac{4}{3}\pi r^3$, so the fractional change in volume is $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$. Therefore, $$\frac{\Delta r}{r} = \frac{1}{3} \frac{\Delta V}{V} = \frac{P}{3B}$$.
Question 8:
moderateThe approximate depth of an ocean is $2700 \text{ m}$. The compressibility of water is $45.4 \times 10^{-11} \text{ Pa}^{-1}$ and density of water is $10^3 \text{ kg/m}^3$. What fractional compression of water will be obtained at the bottom of the ocean?
(2015)
Pressure at depth $h$ is $$P = \rho gh = 10^3 \times 9.8 \times 2700 \approx 26.4 \times 10^6 \text{ Pa}$$. Fractional compression is $$\frac{\Delta V}{V} = P \times K = (26.4 \times 10^6) \times (45.4 \times 10^{-11}) \approx 1.2 \times 10^{-2}$$.
Question 9:
moderateWhen a block of mass $M$ is suspended by a long wire of length $L$, the length of the wire becomes $(L + l)$. The elastic potential energy stored in the extended wire is :
(2019)
The elastic potential energy stored in a stretched wire is given by $U = \frac{1}{2} \times \text{Force} \times \text{Extension}$. Here, the applied force is the weight of the block $Mg$ and the extension is $l$. Therefore, $U = \frac{1}{2} Mgl$.
Question 10:
moderateA rectangular film of liquid is extended from $(4 \text{ cm} \times 2 \text{ cm})$ to $(5 \text{ cm} \times 4 \text{ cm})$. If the work done is $3 \times 10^{-4} \text{ J}$, the value of the surface tension of the liquid is:
(2016 – II)
Work done in stretching a liquid film is $W = T \times 2\Delta A$ (since it has two surfaces). The change in area is $\Delta A = (5 \times 4) - (4 \times 2) = 12 \text{ cm}^2 = 12 \times 10^{-4} \text{ m}^2$. Thus, $$T = \frac{3 \times 10^{-4}}{2 \times 12 \times 10^{-4}} = 0.125 \text{ Nm}^{-1}$$.