Solid and Fluids - NEET Physics Chapterwise MCQs & PYQs
Rankers Physics
One Stop for Your NEET Physics Preparation
NEET Solid and Fluids MCQs & PYQs
Practice NEET Solid and Fluids Questions
Question 31:
moderate
The velocity of a small ball of mass \(M\) and density \(d\), when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is \(\frac{d}{2}\), then the viscous force acting on the ball will be
When the ball reaches terminal velocity, net force is zero: \(F_v + F_B = Mg\). The buoyant force is \(F_B = V \rho_{\text{glycerine}} g = V \left(\frac{d}{2}\right) g = \frac{Mg}{2}\). Thus, the viscous force is \(F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}\).
Assertion (A): Weight of an empty balloon measured in air is \(W_1\). If air at atmospheric pressure is filled inside balloon and again weight of the balloon is measured. Weight of balloon in second case is equal to \(W_1\).
Reason (R): Upthrust is equal to weight of the fluid displaced by the body.
Concept: Archimedes' Principle and apparent weight. When air at atmospheric pressure is filled into a balloon, the weight of the air inside is equal to the upthrust exerted by the surrounding air on the volume displaced by this internal air. Thus, the net change in apparent weight due to the air inside is zero. Both Assertion and Reason are true, and Reason explains Assertion by defining upthrust as per Archimedes' principle.
The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length \(100 \text{cm}\) to stretch it by \(1 \text{mm}\) is (if Young’s modulus of the wire \(= 2.0 \times 10^{11} \text{N} \text{m}^{-2}\)
Energy density is \(u = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} Y \left(\frac{\Delta l}{l}\right)^2 = \frac{1}{2} \times (2.0 \times 10^{11}) \times \left(\frac{10^{-3} \text{m}}{1 \text{m}}\right)^2 = 10^5 \text{J/m}^3\).
The Young’s modulus of wire of length L and radius r is Y. If the length and radius are reduced to \(\frac{L}{3}\) and \(\frac{r}{2}\), then its Young’s modulus will be
Young's modulus is an intrinsic property of the material of the wire and is independent of its geometrical dimensions.
Experimental 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 \[L = \frac{T_B l_A - T_A l_B}{T_B - T_A}\].
Two 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, $F' = \frac{Y (3A) \Delta l}{L_1/3} = 9 \left( \frac{Y A \Delta l}{L_1} \right) = 9F$.
The 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, $$\frac{F_s}{F_b} = \frac{Y_s}{Y_b} = \frac{2}{1} = 2:1$$.
The 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 $$\frac{\Delta V}{V} = P \times K = (26.4 \times 10^6) \times (45.4 \times 10^{-11}) \approx 1.2 \times 10^{-2}$$.
When 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, $U = \frac{1}{2} Mgl$.