Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is ______. (Take surface tension of liquid $= 0.08$ N/m)
(1) $10$
(2) $7$
(3) $8$
(4) $11$
Show answer and solution
Answer: Option (2)
Key idea: Surface Energy and Surface Tension
Given: Initial radius $R = 1$ mm $= 10^{-3}$ m, $n = 512$ droplets, surface tension $T = 0.08$ N/m
Using volume conservation: $R^3 = n \times r^3$, so $r^3 = \frac{R^3}{512} = \frac{R^3}{8^3}$, giving $r = \frac{R}{8} = 0.125 \times 10^{-3}$ m
A uniform wire of length $l$ of weight $w$ is suspended from the roof with a weight of W at the other end. The stress in the wire at $\frac{l}{3}$ distance from the top is $\left(\frac{W}{A} + \frac{2}{\gamma}\frac{w}{A}\right)$, where, $A$ is the cross sectional area of the wire. The value of $\gamma$ is __________.
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Answer: 3
Key idea: Stress distribution in a hanging wire with non-uniform loading
Identify that stress at any point = (Total force below that point)/Area
At distance $\frac{l}{3}$ from top, remaining length below = $l - \frac{l}{3} = \frac{2l}{3}$
Weight per unit length of wire = $\frac{w}{l}$, so weight of wire below = $\frac{w}{l} \times \frac{2l}{3} = \frac{2w}{3}$
Total force at this point = $W + \frac{2w}{3}$ (weight at bottom + wire below)
A tub is filled with water and a wooden cube 10 cm $\times$ 10 cm $\times$ 10 cm is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by 3.87 cm. The mass of the metal coin is __________ gram.
(Take water density as 1 g/cm$^3$ and density of wood as 0.4 g/cm$^3$)
Show answer and solution
Answer: 387
Key idea: Archimedes' Principle and Floatation
Find mass of wooden cube: $m_{wood} = 10 \times 10 \times 10 \times 0.4 = 400$ g
For floating cube without coin, apply equilibrium: $m_{wood} \cdot g = \rho_{water} \cdot g \cdot A \cdot h_1$ where $A = 100$ cm²
The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is $\alpha\pi \times 10^{-4}$ J. The value of $\alpha$ is ________. (Take $\pi = 3.14$)
(1) 0.86
(2) 0.64
(3) 1.92
(4) 7.68
Show answer and solution
Answer: Option (3)
Key idea: Work done in changing surface area of soap bubble (with two surfaces)
Convert diameters to radii: $r_1 = 1$ cm $= 0.01$ m, $r_2 = 3$ cm $= 0.03$ m
For a soap bubble (two surfaces): Total surface area $= 8\pi r^2$
Initial area: $A_1 = 8\pi(0.01)^2 = 8\pi \times 10^{-4}$ m²
Final area: $A_2 = 8\pi(0.03)^2 = 72\pi \times 10^{-4}$ m²
Change in area: $\Delta A = (72-8)\pi \times 10^{-4} = 64\pi \times 10^{-4}$ m²
Work done: $W = T \times \Delta A = 0.03 \times 64\pi \times 10^{-4} = 1.92\pi \times 10^{-4}$ J
Comparing with $\alpha\pi \times 10^{-4}$ J, we get $\alpha = 1.92$
If an air bubble of diameter 2 mm rises steadily through a liquid of density 2000 kg/m$^3$ at a rate of 0.5 cm/s, then the coefficient of viscosity of liquid is ________ Poise. (Take g = 10 m/s$^2$)
(1) 0.88
(2) 8.8
(3)
(4)
Show answer and solution
Answer: Option (2)
Key idea: Stokes' Law and Terminal Velocity of Rising Bubble
Given: diameter $d = 2$ mm, so radius $r = 1$ mm $= 10^{-3}$ m, $\rho = 2000$ kg/m³, $v = 0.5$ cm/s $= 5 \times 10^{-3}$ m/s
For a bubble rising at terminal velocity, use Stokes' law: $v = \frac{2r^2 g \rho}{9\eta}$ (neglecting air density)
Rearrange for viscosity: $\eta = \frac{2r^2 g \rho}{9v}$
The increase in the pressure required to decrease the volume $(\Delta V)$ of water is $6.3 \times 10^7$ N/m$^2$. The percentage decrease in the volume is ______. (Bulk modulus of water = $2.1 \times 10^9$ N/m$^2$.)
(1) $2\%$
(2) $3\%$
(3) $6\%$
Show answer and solution
Answer: Option (2)
Key idea: Bulk Modulus and Volume Change
Write the bulk modulus formula: $B = -\frac{\Delta P}{\Delta V/V}$
A string $A$ of length 0.314 m and Young's modulus $2 \times 10^{10}$ N/m$^2$ is connected to another string $B$ of length and Young's modulus both twice of those of $A$. This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass 0.8 kg. The net change in length of the combination is _______ mm.
(radius of both the strings is 0.2 mm and acceleration due to gravity =10 m/s$^2$)
(Mass of both strings is to be neglected as compared to the mass of load)
(1) 3
(2) 2
(3) 1.9
(4) 1
Show answer and solution
Answer: Option (2)
Key idea: Young's Modulus and Elongation of Strings in Series
Calculate the tension force: $F = mg = 0.8 \times 10 = 8$ N
Calculate elongation of string A using $\Delta L = \frac{FL}{AY}$: $\Delta L_A = \frac{8 \times 0.314}{4\pi \times 10^{-8} \times 2 \times 10^{10}} = 1$ mm
Calculate elongation of string B: $\Delta L_B = \frac{8 \times 0.628}{4\pi \times 10^{-8} \times 4 \times 10^{10}} = 1$ mm
Total elongation: $\Delta L_{total} = \Delta L_A + \Delta L_B = 1 + 1 = 2$ mm
The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _______. $(\pi = 22/7)$
Show answer and solution
Answer: 264
Key idea: Work done in changing surface area of soap bubble with two free surfaces
For a soap bubble, there are two surfaces (inner and outer), so total surface area = $2 \times 4\pi r^2 = 8\pi r^2$
Change in surface area: $\Delta A = 8\pi(r_2^2 - r_1^2) = 8 \times \frac{22}{7} \times (0.02^2 - 0.01^2)$
A metal string $A$ is suspended from a rigid support and its free end is attached to a block of mass $M$. Second block having mass $2M$ is suspended at the bottom of the first block using a string $B$. The area of cross sections of strings $A$ and $B$ are same. The ratio of lengths of strings of $A$ to $B$ is $2$ and the ratio of their Young's moduli $(Y_A / Y_B)$ is $0.5$. The ratio of elongations in $A$ to $B$ is ________.
(1) 1
(2) 4
(3)
(4)
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Young's Modulus and Elongation of Strings
A water spray gun is attached to a hose of cross sectional area 30 cm$^2$. The gun comprises of 10 perforations each of cross sectional area of 15 mm$^2$. If the water flows in the hose with the speed of 50 cm/s, calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)
(1) 100 m/s
(2) 10 m/s
(3) 1000 m/s
(4) $15 \times 10^2$ m/s
Show answer and solution
Answer: Option (2)
Key idea: Continuity Equation (Equation of Continuity)
Write the continuity equation: $A_1 v_1 = n \times a \times v_2$ where $n$ is the number of perforations
Convert all areas to consistent units: $A_1 = 30$ cm², $a = 15$ mm² $= 0.15$ cm²
The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $T_1$ to $T_2$ resulting in increase of its length by $\Delta L_1$. The temperature is further increased from $T_2$ to $T_3$ such that the increase in its length is $\Delta L_2$.
Given $T_3 - T_1 = 2T_2$ and $T_2 - T_1 = \Delta T$, the value of $\Delta L_2$ is ______.
(1) $\Delta L_1[1 + 2\alpha^2 (\Delta T)^2]$
Show answer and solution
Answer: Option (1)
Key idea: Thermal expansion with successive temperature changes
1. For first expansion: $\Delta L_1 = L_0 \alpha (T_2 - T_1) = L_0 \alpha \Delta T$, so $L_0 = \frac{\Delta L_1}{\alpha \Delta T}$
2. After first expansion, new length: $L_1 = L_0(1 + \alpha \Delta T)$
Two wires as shown in the figure below. made of steel and have breaking stress of $12 \times 10^8$ N/m$^2$. Area of cross-section of upper wire is 0.008 cm$^2$ and of lower wire is 0.004 cm$^2$. The maximum mass that can be added to pan without breaking any wire is ______ kg.
(take $g = 10$ m/s$^2$)
[Figure: A vertical arrangement showing a fixed support at top, connected to upper wire, then a 30 kg mass, then lower wire, then a 10 kg mass, and finally a pan at the bottom]
(1) 56
(2) 38
(3) 96
(4) 56
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
A cube has side length 5 cm and modulus of rigidity $10^5$ N/m$^2$. The displacement produced by a force of 10 N in the upper face of cube is _____ mm.
Show answer and solution
Answer: 2
Key idea: Shear Modulus and Shear Deformation
Write the formula for modulus of rigidity: $G = \frac{F/A}{\Delta x/L}$ where $F$ is tangential force, $A$ is area, $\Delta x$ is displacement, and $L$ is height
Rearrange to find displacement: $\Delta x = \frac{F \times L}{A \times G}$
Calculate area of cube face: $A = (0.05)^2 = 0.0025$ m²
Substitute values: $\Delta x = \frac{10 \times 0.05}{0.0025 \times 10^5}$
Simplify: $\Delta x = \frac{0.5}{250} = 0.002$ m = 2 mm
Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is __________. (S is the surface tension of mercury).
(1) $8\pi r^2 S$
(2) $16\pi r^2 S$
(3) $64\pi r^2 S$
(4) $4\pi r^2 S$
Show answer and solution
Answer: Option (2)
Key idea: Surface Energy and Coalescence of Drops
Use volume conservation: $8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$, which gives $R^3 = 8r^3$, so $R = 2r$
A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is $600 \times 10^{-6}$ m$^3$. The elastic potential energy stored in the wire in stretched condition would be __________ J.
(Given Young modulus of copper = $1.1 \times 10^{11}$ N/m$^2$)
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Answer: 33
Key idea: Elastic Potential Energy in Stretched Wire
Given: $L = 3$ m, $\Delta L = 3 \times 10^{-3}$ m, $V = 600 \times 10^{-6}$ m³, $Y = 1.1 \times 10^{11}$ N/m²
The heat extracted out of $x$ gram of water initially at 50 °C to cool it down to 0 °C is sufficient to evaporate $(1000 - x)$ gram of water also initially at 50 °C. The value of $x$ (closest integer) is __________.
(Take latent heat of water 2256 kJ/kg. K, specific heat capacity of water 4200 J/kg. K)
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Answer: 922
Key idea: Heat Transfer and Phase Change - Energy Conservation
Heat extracted from $x$ g water cooling from 50°C to 0°C: $Q_1 = \frac{x}{1000} \times 4200 \times 50 = 210x$ J
Heat needed to raise $(1000-x)$ g water from 50°C to 100°C: $Q_{2a} = \frac{1000-x}{1000} \times 4200 \times 50 = 210(1000-x)$ J
Heat needed to evaporate $(1000-x)$ g water at 100°C: $Q_{2b} = \frac{1000-x}{1000} \times 2256000 = 2256(1000-x)$ J
Figure represents the extension ($\Delta l$) of a wire of length 1 meter, suspended from the ceiling of the room at one end with a load W connected to the other end. If the cross-sectional area of the wire is $10^{-5}$ m$^2$ then the Young's modulus of the wire is __________ N/m$^2$. [Figure: A graph showing $\Delta l (\times 10^{-4}$ m) on y-axis ranging from 0 to 6, and W (N) on x-axis ranging from 0 to 60. A straight line passes through origin with positive slope, showing linear relationship between extension and load.]
(1) $1.0 \times 10^{11}$
(2)
(3)
(4)
Show answer and solution
Answer: Option (1)
Key idea: Young's Modulus and Stress-Strain Relationship
Read from graph: At $W = 60$ N, $\Delta l = 0.6 \times 10^{-4}$ m
A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is 528 dm$^3$ (dm : decimeter) The vessel is placed on a solid block of exactly same height as vessel. If a small hole is made at 70 cm below the top of water level, then horizontal range of water falling on the ground in the beginning is ________ cm.
(1) $120\sqrt{2}$
(2) $140\sqrt{2}$
(3) $140\sqrt{3}$
(4) $120\sqrt{3}$
Show answer and solution
Answer: Option (2)
Key idea: Torricelli's Theorem and Projectile Motion
Find height of vessel: $V = \pi r^2 h \Rightarrow h = \frac{528 \times 10^{-3}}{\pi (0.4)^2} = 1.05$ m = 105 cm
Hole is 70 cm below water surface. Height of hole above ground = (105 + 105) - 70 = 140 cm
Velocity of efflux using Torricelli's theorem: $v = \sqrt{2gh_1} = \sqrt{2 \times 10 \times 0.7} = \sqrt{14}$ m/s
Time to fall from height $h_2 = 140$ cm: $t = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 \times 1.4}{10}} = \sqrt{0.28} = \frac{\sqrt{7}}{5}$ s
Horizontal range: $R = vt = \sqrt{14} \times \frac{\sqrt{7}}{5} = \frac{\sqrt{98}}{5} = \frac{7\sqrt{2}}{5}$ m = $140\sqrt{2}$ cm
A liquid of density 600 kg/m$^3$ flowing steadily in a tube of varying cross-section. The cross-section at a point $A$ is 1.0 cm$^2$ and that at $B$ is 20 mm$^2$. Both the points $A$ and $B$ are in same horizontal plane, the speed of the liquid at $A$ is 10 cm/s. The difference in pressures at $A$ and $B$ points is _________ Pa.
(1) 18
(2) 144
(3) 36
(4) 72
Show answer and solution
Answer: Option (4)
Key idea: Continuity Equation and Bernoulli's Principle for fluid flow in varying cross-sections
Convert all units: $A_A = 1.0 imes 10^{-4}$ m², $A_B = 2.0 imes 10^{-5}$ m², $v_A = 0.10$ m/s
A spherical liquid drop of radius $R$ acquires the terminal velocity $v_1$ when falls through a gas of viscosity $\eta$. Now the drop is broken into 64 identical droplets and each droplet acquires terminal velocity $v_2$ falling through the same gas. The ratio of terminal velocities $v_1/v_2$ is ______.
(1) 4
(2) 0.25
(3) 32
(4) 16
Show answer and solution
Answer: Option (4)
Key idea: Terminal velocity and Stokes' law - terminal velocity is proportional to the square of the radius of a spherical object falling through a viscous medium
For a sphere falling through viscous medium, terminal velocity $v = \frac{2r^2\rho g}{9\eta}$, so $v \propto r^2$
Apply volume conservation: Original drop volume = 64 × (small droplet volume)
