Question 101:
easyEight drops of equal radii are falling through air with a steady velocity of \(3\text{ cm/s}\). If the eight drops combine to form a single drop, then its steady velocity will be
Terminal velocity \(v \propto r^2\). Since volume remains constant, \(\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 \implies R = 2r\). Thus, the new terminal velocity is \(v' = \left(\frac{R}{r}\right)^2 v = 2^2 \times 3 = 12\text{ cm/s}\).
Question 102:
easyThe experiment which is used to determine Young’s modulus of the material of a given wire, is
Searle's apparatus/experiment is specifically designed and used to determine the Young's modulus of elasticity of a metal wire by measuring elongation under load.
Question 103:
moderateExperimental observations show that for given solid material, the magnitude of strain produced is same whether the stress is tensile or compressive. The ratio of tensile stress to longitudinal strain is defined as Young’s modulus and is denoted by \(Y = \sigma/e\). The length of a metal wire is \(l_A\) when the tension in it is \(T_A\) and is \(l_B\) when tension is \(T_B\). The natural length of wire is
Let the natural length be \(L\). Using Hooke's law, \[l_A = L(1 + T_A/AY)\] and \[l_B = L(1 + T_B/AY)\]. Eliminating \(AY\) gives \[L = \frac{T_B l_A - T_A l_B}{T_B - T_A}\].
Question 104:
easyA U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with \(12\text{ cm}\) of water in one arm and \(15\text{ cm}\) of spirit in other. The specific gravity of spirit is
At the interface level, the pressure on both sides must be equal: \(h_w \rho_w g = h_s \rho_s g\). Thus, \(\rho_s / \rho_w = h_w / h_s = 12 / 15 = 0.8\).
Question 105:
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 106:
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 107:
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 108:
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 109:
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 110:
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 $Y = \frac{MgL}{A(L_1 - L)}$.