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Multiple Choice Questions

One and More than One Correct

1. A student performs an experiment for determination of $g \left(= \frac{4\pi^2 l}{T^2}\right)$, $l \approx 1\text{ m}$, and he commits an error of $\Delta l$. For $T$ he takes the time of $n$ oscillations with the stop watch of least count $\Delta T$ and he commits a human error of $0.1\text{ sec}$. For which of the following data, the measurement of $g$ will be most accurate? [JEE 2006]

a. $\Delta l = 5\text{ mm}, \Delta T = 0.2\text{ sec}, n = 10, \text{Amplitude} = 5\text{ mm}$

b. $\Delta l = 5\text{ mm}, \Delta T = 0.2\text{ sec}, n = 20, \text{Amplitude} = 5\text{ mm}$

c. $\Delta l = 5\text{ mm}, \Delta T = 0.1\text{ sec}, n = 20, \text{Amplitude} = 1\text{ mm}$

d. $\Delta l = 1\text{ mm}, \Delta T = 0.1\text{ sec}, n = 50, \text{Amplitude} = 1\text{ mm}$

Answer: (d)

Solution:

For measurement of $g$ to be most accurate, the fractional error $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$ must be minimum. Here $\Delta l$ is minimum ($1\text{ mm}$), number of oscillations $n$ is maximum ($50$), and amplitude of oscillation is small ($1\text{ mm}$), giving the least total error.

2. The circular divisions of shown screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball is [JEE 2006]

a. 2.25 mm

b. 2.20 mm

c. 1.20 mm

d. 1.25 mm

Answer: (c)

Solution:

Least Count (L.C.) $= \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}$

Zero error $= 5 \times 0.01\text{ mm} = 0.05\text{ mm}$

Measured value $= 0.5\text{ mm} \times 2 + 25 \times 0.01\text{ mm} - 0.05\text{ mm} = 1\text{ mm} + 0.25\text{ mm} - 0.05\text{ mm} = 1.20\text{ mm}$

3. A student performs an experiment to determine the Young’s modulus of wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of $\pm 0.05\text{ mm}$ at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of $\pm 0.01\text{ mm}$. Take $g = 9.8\text{ m/s}^2$ (exact). The Young’s modulus obtained from the reading is: [JEE 2007 P-II]

a. $(2.0 \pm 0.3) \times 10^{11}\text{ N/m}^2$

b. $(2.0 \pm 0.2) \times 10^{11}\text{ N/m}^2$

c. $(2.0 \pm 0.1) \times 10^{11}\text{ N/m}^2$

d. $(2.0 \pm 0.05) \times 10^{11}\text{ N/m}^2$

Answer: (b)

Solution:

$$Y = \frac{4FL}{\pi d^2 \Delta L}$$ $$\frac{\Delta Y}{Y} = 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta(\Delta L)}{\Delta L} = 2\left(\frac{0.01}{0.4}\right) + \frac{0.05}{0.8} = 0.05 + 0.0625 = 0.1125$$ $$\Delta Y = 2.0 \times 10^{11} \times 0.1125 \approx 0.2 \times 10^{11}\text{ N/m}^2$$ $$Y = (2.0 \pm 0.2) \times 10^{11}\text{ N/m}^2$$

4. A Vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier calipers, the least count is [JEE 2010 P-II]

a. 0.02 mm

b. 0.05 mm

c. 0.1 mm

d. 0.2 mm

Answer: (d)

Solution:

$$\text{L.C.} = 1\text{ M.S.D.} - 1\text{ V.S.D.}$$ $$1\text{ V.S.D.} = \frac{16}{20}\text{ M.S.D.} = \frac{4}{5}\text{ mm}$$ $$\text{L.C.} = \left(1 - \frac{4}{5}\right)\text{ mm} = \frac{1}{5}\text{ mm} = 0.2\text{ mm}$$

5. A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true? [JEE 2010 P-I]

a. Error $\Delta T$ in measuring T, the time period, is 0.05 seconds

b. Error $\Delta T$ in measuring T, the time period, is 1 second

c. Percentage error in the determination of g is 5%

d. Percentage error in the determination of g is 2.5%

Answer: (a, c)

Solution:

$$\Delta T = \frac{\Delta t}{n} = \frac{1}{20} = 0.05\text{ sec}$$ $$g = \frac{4\pi^2 L n^2}{t^2}$$ $$\frac{\Delta g}{g} = 2 \frac{\Delta t}{t}$$ $$\%\text{ error} = 2 \times \frac{1}{40} \times 100 = 5\%$$

6. The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is [JEE 2011 P-II]

a. 0.9%

b. 2.4%

c. 3.1%

d. 4.2%

Answer: (c)

Solution:

$$\text{Diameter} = 2.5 + 20 \times \frac{0.5}{50} = 2.70\text{ mm}$$ $$\%\text{ error in density} = \frac{\Delta m}{m} \times 100 + 3 \left(\frac{\Delta d}{d}\right) \times 100$$ $$= 2\% + 3 \times \frac{0.01}{2.70} \times 100 = 3.1\%$$

7. The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is [JEE Adv. 2013 P-I]

a. 5.112 cm

b. 5.124 cm

c. 5.136 cm

d. 5.148 cm

Answer: (b)

Solution:

$$\text{Main scale division (s)} = 0.05\text{ cm}$$ $$\text{Vernier scale division (v)} = \frac{2.45}{50} = 0.049\text{ cm}$$ $$\text{Least count} = 0.05 - 0.049 = 0.001\text{ cm}$$ $$\text{Diameter} = 5.10 + 24 \times 0.001 = 5.124\text{ cm}$$

8. Let $[\varepsilon_0]$ denotes the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = Time and A = electric current, then [JEE Main 2013]

a. $[\varepsilon_0] = [\text{M}^{-1}\text{L}^{-3}\text{T}^2\text{A}]$

b. $[\varepsilon_0] = [\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2]$

c. $[\varepsilon_0] = [\text{M}^{-1}\text{L}^2\text{T}^{-1}\text{A}^{-2}]$

d. $[\varepsilon_0] = [\text{M}^{-1}\text{L}^2\text{T}^{-1}\text{A}]$

Answer: (b)

Solution:

$$[\varepsilon_0] = \left[\frac{\text{C}^2}{\text{N}\cdot\text{m}^2}\right] = \frac{\text{A}^2\text{T}^2}{\text{M}\text{L}\text{T}^{-2} \cdot \text{L}^2} = [\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2]$$

9. Using the expression $2d \sin\theta = \lambda$, one calculates the values of d by measuring the corresponding angles $\theta$ in the range $\theta$ to $90^\circ$. The wavelength $\lambda$ is exactly known and the error in $\theta$ is constant for all values of $\theta$. As $\theta$ increases from $0^\circ$, [JEE Adv. 2013 P-II]

a. the absolute error in d remains constant

b. the absolute error in d increases

c. the fractional error in d remains constant

d. the fractional error in d decreases

Answer: (d)

Solution:

$$d = \frac{\lambda}{2\sin\theta}$$ $$\ln d = \ln\left(\frac{\lambda}{2}\right) - \ln(\sin\theta)$$ $$\frac{\Delta d}{d} = -\cot\theta \Delta\theta \implies \left|\frac{\Delta d}{d}\right| = \cot\theta \Delta\theta$$ $$\text{Also, } \Delta d = d \cot\theta \Delta\theta = \frac{\lambda}{2\sin\theta} \cot\theta \Delta\theta = \frac{\lambda \cos\theta}{2\sin^2\theta} \Delta\theta$$

As $\theta$ increases, $\cot\theta$ decreases and $\frac{\cos\theta}{\sin^2\theta}$ also decreases, so fractional error in $d$ decreases.

10. The current voltage relation of diode is given by $I = (e^{1000V/T} - 1)\text{ mA}$, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring $\pm 0.01\text{V}$ while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA? [JEE Main 2014]

a. 0.5 mA

b. 0.05 mA

c. 0.2 mA

d. 0.02 mA

Answer: (c)

Solution:

$$I = e^{1000V/T} - 1$$ $$dI = e^{1000V/T} \cdot \frac{1000}{T} dV$$ $$I = 5\text{ mA} \implies e^{1000V/T} = 6\text{ mA}$$ $$dI = 6 \left(\frac{1000}{300}\right) (0.01) = 6 \times \frac{10}{3} \times 0.01 = 0.2\text{ mA}$$

11. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? [JEE Main 2014]

a. A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

b. A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.

c. A meter scale.

d. A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale & main scale has 10 divisions in 1 cm.

Answer: (d)

Solution:

Least count $= 0.01\text{ cm} = 0.1\text{ mm}$.

For vernier calliper of 10 V.S.D. $= 9$ M.S.D.

$$\text{L.C.} = 1\text{ M.S.D.} - 1\text{ V.S.D.} = 1\text{ M.S.D.} - 0.9\text{ M.S.D.} = 0.1\text{ M.S.D.} = 0.01\text{ cm}$$

12. Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then: [JEE Adv. 2015 P-I]

a. If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.

b. If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

c. If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.

d. If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

Answer: (b, c)

Solution:

For vernier callipers,

$$\text{1 main scale division} = \frac{1}{8}\text{ cm}$$ $$\text{1 vernier scale division} = \frac{4}{5} \times \frac{1}{8} = \frac{1}{10}\text{ cm}$$ $$\text{Least count} = \frac{1}{8} - \frac{1}{10} = \frac{1}{40}\text{ cm} = \frac{1}{4}\text{ mm} = 0.25\text{ mm}$$

For screw gauge, pitch $= 2 \times \text{main scale division}$.

So, options (b) and (c) are correct.

13. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1s, then the reported mean time should be: [JEE Main 2016]

a. $92 \pm 2\text{ s}$

b. $92 \pm 5.0\text{ s}$

c. $92 \pm 1.8\text{ s}$

d. $92 \pm 3\text{ s}$

Answer: (a)

Solution:

$$\bar{t} = \frac{90 + 91 + 95 + 92}{4} = 92\text{ sec}$$ $$\text{Absolute errors in each reading} = 2, 1, 3, 0$$ $$\text{Mean error} = \frac{2 + 1 + 3 + 0}{4} = 1.5\text{ sec}$$

Since the least count of the measuring clock is 1 second, we round off 1.5 seconds to 2 seconds.

$$\text{So, reported mean time } t = 92 \pm 2\text{ sec}$$

14. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line? [JEE Main 2016]

a. 0.75 mm

b. 0.80 mm

c. 0.70 mm

d. 0.50 mm

Answer: (b)

Solution:

$$\text{L.C.} = \frac{0.5}{50} = 0.01\text{ mm}$$ $$\text{Zero error} = -0.5\text{ mm} + (45)(0.01) = -0.05\text{ mm}$$ $$\text{Measured thickness} = 0.5\text{ mm} + (25)(0.01) = 0.75\text{ mm}$$ $$\text{Actual thickness} = 0.75 - (-0.05)\text{ mm} = 0.80\text{ mm}$$

15. There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers ($C_1$) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper ($C_2$) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers $C_1$ and $C_2$, respectively, are: [JEE Adv. 2016 P-II]

a. 2.87 and 2.87

b. 2.87 and 2.86

c. 2.87 and 2.83

d. 2.85 and 2.82

Answer: (c)

Solution:

For vernier $C_1$:

$$10\text{ VSD} = 9\text{ MSD} = 9\text{ mm} \implies 1\text{ VSD} = 0.9\text{ mm}$$ $$\text{LC} = 1\text{ MSD} - 1\text{ VSD} = 1\text{ mm} - 0.9\text{ mm} = 0.1\text{ mm}$$ $$\text{Reading of } C_1 = \text{MSR} + (\text{VSR})(\text{L.C.}) = 28\text{ mm} + (7)(0.1) = 28.7\text{ mm} = 2.87\text{ cm}$$

For vernier $C_2$:

$$\text{Distance measured from main scale} = \text{distance measured from vernier scale}$$ $$28\text{ mm} + (1\text{ mm})(8) = (28\text{ mm} + x) + (1.1\text{ mm})(7)$$ $$\text{Solving, } x = 0.3\text{ mm}$$ $$\text{Reading of } C_2 = 28\text{ mm} + 0.3\text{ mm} = 2.83\text{ cm}$$

16. A length–scale ($l$) depends on the permittivity ($\varepsilon$) of a dielectric material, Boltzmann constant ($k_B$), the absolute temperature ($T$), the number per unit volume ($n$) of certain charged particles, and the charge ($q$) carried by each of the particles. Which of the following expression(s) for $l$ is (are) dimensionally correct? [JEE Adv. 2016 P-I]

a. $l = \sqrt{\frac{n q^2}{\varepsilon k_B T}}$

b. $l = \sqrt{\frac{\varepsilon k_B T}{n q^2}}$

c. $l = \left(\frac{q^2}{\varepsilon n^{2/3} k_B T}\right)$

d. $l = \left(\frac{q^2}{\varepsilon n^{1/3} k_B T}\right)$

Answer: (b, d)

Solution:

$$\left[\frac{q^2}{\varepsilon k_B T}\right] = \left[\frac{F r^2}{F r}\right] = [L]$$

Checking option (b):

$$\left[\sqrt{\frac{\varepsilon k_B T}{n q^2}}\right] = \left[\sqrt{\frac{1}{L^{-3} \cdot L}}\right] = [L]$$

Checking option (d):

$$\left[\frac{q^2}{\varepsilon n^{1/3} k_B T}\right] = \left[\frac{L}{L^{-1}}\right] = [L]$$

Correct options are B and D.

17. In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a period of a periodic motion is $T = 2\pi \sqrt{\frac{7(R-r)}{5g}}$. The values of R and r are measured to be $(60 \pm 1)\text{ mm}$ and $(10 \pm 1)\text{ mm}$ respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is (are) true? [JEE Adv. 2016 P-II]

a. The error in the measurement of r is 10%

b. The error in the measurement of T is 3.57%

c. The error in the measurement of T is 2%

d. The error in the determined value of g is 11%

Answer: (a, b, d)

Solution:

$\bar{T} = \frac{0.52 + 0.56 + 0.57 + 0.54 + 0.59}{5} = 0.56\text{ s}$

$\Delta T_{\text{mean}} = \frac{0.04 + 0.00 + 0.01 + 0.02 + 0.03}{5} = 0.02\text{ s}$

$$\%\text{ error in } T = \frac{0.02}{0.56} \times 100 = 3.57\%$$ $$\%\text{ error in } r = \frac{1}{10} \times 100 = 10\%$$ $$T^2 \propto \frac{R-r}{g} \implies \frac{\Delta g}{g} = 2 \frac{\Delta T}{T} + \frac{\Delta R + \Delta r}{R - r}$$ $$\frac{\Delta g}{g} = 2(3.57\%) + \frac{1 + 1}{60 - 10} \times 100\% = 7.14\% + 4\% = 11.14\% \approx 11\%$$

18. A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is $\delta T = 0.01\text{ seconds}$ and he measures the depth of the well to be $L = 20\text{ meters}$. Take the acceleration due to gravity $g = 10\text{ ms}^{-2}$ and the velocity of sound is $300\text{ ms}^{-1}$. Then the fractional error in the measurement, $\delta L/L$, is closest to [JEE Adv. 2017 P-II]

a. 1%

b. 5%

c. 3%

d. 0.2%

Answer: (a)

Solution:

$$T = t_1 + t_2 = \sqrt{\frac{2L}{g}} + \frac{L}{V}$$ $$\Delta T = \frac{1}{2}\sqrt{\frac{2}{g L}} \Delta L + \frac{\Delta L}{V}$$ $$0.01 = \left(\frac{1}{2\sqrt{100}} + \frac{1}{300}\right) \Delta L = \left(\frac{1}{20} + \frac{1}{300}\right) \Delta L = \frac{16}{300} \Delta L$$ $$\Delta L = \frac{3}{16}$$ $$\frac{\Delta L}{L} = \frac{3}{16 \times 20} = \frac{3}{320} \approx 1\%$$

19. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is: [JEE Main 2018]

a. 6%

b. 2.5%

c. 3.5%

d. 4.5%

Answer: (d)

Solution:

$$\rho = \frac{m}{l^3}$$ $$\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 3\frac{\Delta l}{l} = 1.5\% + 3(1\%) = 4.5\%$$

20. Dimension of $\sqrt{\frac{\varepsilon_0}{\mu_0}}$ are: [JEE Main 2019]

a. $[\text{M}^{-2}\text{L}^{-3}\text{T}^2\text{A}^2]$

b. $[\text{M}^{-1}\text{L}^{-2}\text{T}^3\text{A}^2]$

c. $[\text{M}^{-2}\text{L}^2\text{T}^{-3}\text{A}^2]$

d. $[\text{M}^2\text{L}^2\text{T}^{-3}\text{A}^2]$

Answer: (b)

Solution:

$$\sqrt{\frac{\varepsilon_0}{\mu_0}} = \varepsilon_0 c = [\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2][\text{L}\text{T}^{-1}] = [\text{M}^{-1}\text{L}^{-2}\text{T}^3\text{A}^2]$$

21. Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to; [JEE Main 2019]

a. $\sqrt{\frac{Gh}{c^3}}$

b. $\sqrt{\frac{hc}{G^5}}$

c. $\sqrt{\frac{c^3}{Gh}}$

d. $\sqrt{\frac{Gh}{c^5}}$

Answer: (d)

Solution:

$$t \propto G^x h^y c^z$$ $$[\text{T}] = [\text{M}^{-1}\text{L}^3\text{T}^{-2}]^x [\text{M}\text{L}^2\text{T}^{-1}]^y [\text{L}\text{T}^{-1}]^z$$ $$-x + y = 0 \implies x = y$$ $$3x + 2y + z = 0 \implies 5x + z = 0$$ $$-2x - y - z = 1 \implies -3x - z = 1$$ $$\text{Solving gives } x = y = \frac{1}{2}, z = -\frac{5}{2} \implies t \propto \sqrt{\frac{Gh}{c^5}}$$

22. The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 division below the mean line. The readings of the main scale and the circular scales, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is: [JEE Main 2019]

a. 5.755 mm

b. 5.725 mm

c. 5.740 mm

d. 5.95 mm

Answer: (b)

Solution:

$$\text{L.C.} = \frac{0.5}{100} = 0.005\text{ mm}$$ $$\text{Positive error} = +3 \times 0.005 = 0.015\text{ mm}$$ $$\text{Thickness} = 5.5\text{ mm} + (48 \times 0.005\text{ mm}) - 0.015\text{ mm} = 5.5 + 0.24 - 0.015 = 5.725\text{ mm}$$

23. The diameter and height of a cylinder are measured by a meter scale to be $12.6 \pm 0.1\text{ cm}$ and $34.2 \pm 0.1\text{ cm}$, respectively. What will be the value of its volume in appropriate significant figures? [JEE Main 2019]

a. $4260 \pm 80\text{ cm}^3$

b. $4300 \pm 80\text{ cm}^3$

c. $4264.4 \pm 81.0\text{ cm}^3$

d. $4264 \pm 81\text{ cm}^3$

Answer: (a)

Solution:

$$V = \frac{\pi d^2 h}{4} = \frac{\pi (12.6)^2 (34.2)}{4} = 4264.4\text{ cm}^3 \approx 4260\text{ cm}^3$$ $$\frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta h}{h} = 2\left(\frac{0.1}{12.6}\right) + \frac{0.1}{34.2} = 0.0188$$ $$\Delta V = 4264.4 \times 0.0188 = 80.1\text{ cm}^3 \approx 80\text{ cm}^3$$

24. The density of a material in SI units is $128\text{ kg m}^{-3}$. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is: [JEE Main 2019]

a. 410

b. 640

c. 16

d. 40

Answer: (d)

Solution:

$$n_1 u_1 = n_2 u_2$$ $$128 \frac{\text{kg}}{\text{m}^3} = n_2 \frac{50\text{ g}}{(25\text{ cm})^3}$$ $$n_2 = 128 \times \left(\frac{1000\text{ g}}{50\text{ g}}\right) \times \left(\frac{25\text{ cm}}{100\text{ cm}}\right)^3 = 128 \times 20 \times \frac{1}{64} = 40$$

25. If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be: [JEE Main 2019]

a. V$^{-2}$ A$^2$ F$^2$

b. V$^{-4}$ A$^2$ F

c. V$^{-4}$ A$^{-2}$ F

d. V$^{-2}$ A$^2$ F$^{-2}$

Answer: (b)

Solution:

$$[Y] = \frac{[F]}{[A_{\text{area}}]} = [\text{F}][\text{L}]^{-2}$$ $$\text{Since } [\text{V}] = [\text{L}][\text{T}]^{-1} \text{ and } [\text{A}] = [\text{L}][\text{T}]^{-2} \implies [\text{L}] = \frac{[\text{V}]^2}{[\text{A}]}$$ $$[Y] = [\text{F}] \left(\frac{[\text{V}]^2}{[\text{A}]}\right)^{-2} = [\text{V}]^{-4} [\text{A}]^2 [\text{F}]$$

26. The force of interaction between two atoms is given by $F = \alpha \beta \exp\left(-\frac{x^2}{\alpha k T}\right)$; where x is the distance, k is the Boltzmann constant and T is temperature and $\alpha$ and $\beta$ are two constants. The dimension of $\beta$ is: [JEE Main 2019]

a. M$^2$ L$^2$ T$^{-2}$

b. M$^2$ L T$^{-4}$

c. M$^0$ L$^2$ T$^{-4}$

d. M L T$^{-2}$

Answer: (b)

Solution:

$$\left[\frac{x^2}{\alpha k T}\right] = 1 \implies [\alpha] = \frac{[\text{L}^2]}{[\text{M}\text{L}^2\text{T}^{-2}]} = [\text{M}^{-1}\text{T}^2]$$ $$[F] = [\alpha][\beta] \implies [\text{M}\text{L}\text{T}^{-2}] = [\text{M}^{-1}\text{T}^2][\beta]$$ $$[\beta] = [\text{M}^2\text{L}\text{T}^{-4}]$$

27. Let $L$, $r$, $C$ and $V$ represent inductance, resistance, capacitance and voltage, respectively. The dimension of $\frac{r C V}{L}$ in SI units will be: [JEE Main 2019]

a. [LTA]

b. [LA$^{-2}$]

c. [A$^{-1}$]

d. [LT$^2$]

Answer: (c)

Solution:

$$\left[\frac{L}{r}\right] = [\text{T}] \quad \text{and} \quad [CV] = [Q] = [\text{A}\text{T}]$$ $$\left[\frac{r C V}{L}\right] = \frac{[\text{A}\text{T}]}{[\text{T}]} = [\text{A}]$$

Wait, for $\frac{r C V}{L}$, we have $\frac{C V}{L/r} = \frac{\text{A T}}{\text{T}} = \text{A}$. Thus $\left[\frac{L}{r C V}\right] = [\text{A}^{-1}]$. So dimension is $[\text{A}^{-1}]$.

28. The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure $5\mu\text{m}$ diameter of wire is: [JEE Main 2019]

a. 50

b. 100

c. 200

d. 500

Answer: (c)

Solution:

$$\text{Least Count} \le 5 \times 10^{-6}\text{ m} = 5 \times 10^{-3}\text{ mm}$$ $$\text{Least Count} = \frac{\text{Pitch}}{N} = \frac{1\text{ mm}}{N} \le 5 \times 10^{-3}\text{ mm}$$ $$N \ge \frac{1}{5 \times 10^{-3}} = 200$$

29. If in a hypothetical system if the angular momentum and mass are dimensionless. Then which of the following is true. [JEE Adv. 2019 P-I]

a. The linear momentum varies as L$^{-1}$

b. The energy varies as L$^{-2}$

c. The power varies as L$^{-4}$

d. The force varies as L$^{-5}$

Answer: (a, b, c)

Solution:

$$[\text{M}] = [1]$$ $$[J] = [\text{M}\text{L}^2\text{T}^{-1}] = [1] \implies [\text{T}] = [\text{L}^2]$$ $$\text{Linear momentum } [p] = [\text{M}\text{L}\text{T}^{-1}] = [\text{L}\text{L}^{-2}] = [\text{L}^{-1}]$$ $$\text{Energy } [E] = [\text{M}\text{L}^2\text{T}^{-2}] = [\text{L}^2\text{L}^{-4}] = [\text{L}^{-2}]$$ $$\text{Power } [P] = [\text{M}\text{L}^2\text{T}^{-3}] = [\text{L}^2\text{L}^{-6}] = [\text{L}^{-4}]$$ $$\text{Force } [F] = [\text{M}\text{L}\text{T}^{-2}] = [\text{L}\cdot\text{L}^{-4}] = [\text{L}^{-3}]$$

Matrix Match/Column Match

30. Match Column I with Column II and select the correct answer using the codes given below the lists:

Column I

(A) Boltzmann Constant

(B) Coefficient of viscosity

(D) Thermal conductivity

Column II

1. $[\text{M}\text{L}^2\text{T}^{-1}]$

2. $[\text{M}\text{L}^{-1}\text{T}^{-1}]$

3. $[\text{M}\text{L}\text{T}^{-3}\text{K}^{-1}]$

4. $[\text{M}\text{L}^2\text{T}^{-2}\text{K}^{-1}]$

a. A$\rightarrow$4; B$\rightarrow$2; C$\rightarrow$1; D$\rightarrow$3

b. A$\rightarrow$1; B$\rightarrow$3; C$\rightarrow$4; D$\rightarrow$2

c. A$\rightarrow$2; B$\rightarrow$1; C$\rightarrow$3; D$\rightarrow$4

d. A$\rightarrow$3; B$\rightarrow$4; C$\rightarrow$2; D$\rightarrow$1

Answer: (a)

Solution:

(A) Boltzmann constant $k = \frac{\text{Energy}}{\text{Temperature}} \implies [\text{M}\text{L}^2\text{T}^{-2}\text{K}^{-1}]$

(B) Coefficient of viscosity $\eta \implies [\text{M}\text{L}^{-1}\text{T}^{-1}]$

(D) Thermal conductivity $K \implies [\text{M}\text{L}\text{T}^{-3}\text{K}^{-1}]$

Paragraphs

31. The relation between $[\varepsilon_0]$ and $[\mu_0]$ is:

a. $[\mu_0] = [\varepsilon_0][\text{L}]^2[\text{T}]^{-2}$

b. $[\mu_0] = [\varepsilon_0][\text{L}]^{-2}[\text{T}]^2$

c. $[\mu_0]^{-1} = [\varepsilon_0][\text{L}]^2[\text{T}]^{-2}$

d. $[\mu_0]^{-1} = [\varepsilon_0][\text{L}]^{-2}[\text{T}]^{-2}$

Answer: (d)

Solution:

$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \implies \mu_0 \varepsilon_0 = \frac{1}{c^2}$$ $$[\mu_0]^{-1} = [\varepsilon_0][c]^2 = [\varepsilon_0][\text{L}]^2[\text{T}]^{-2} \implies [\mu_0]^{-1} = [\varepsilon_0][\text{L}]^2[\text{T}]^{-2}$$

Rearranging: $[\mu_0]^{-1} [\varepsilon_0]^{-1} = [\text{L}]^{-2}[\text{T}]^2$, hence $[\mu_0]^{-1} = [\varepsilon_0][\text{L}]^{-2}[\text{T}]^{-2}$.

32. Consider the ratio $r = \frac{1-a}{1+a}$ to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is $\Delta a$ ($\Delta a/a \ll 1$), then what is the error $\Delta r$ in determining r?

a. $\frac{\Delta a}{(1+a)^2}$

b. $\frac{2\Delta a}{(1+a)^2}$

c. $\frac{2\Delta a}{(1-a)^2}$

d. $\frac{2a\Delta a}{(1-a)^2}$

Answer: (b)

Solution:

$$\ln r = \ln(1-a) - \ln(1+a)$$ $$\frac{\Delta r}{r} = \left|-\frac{\Delta a}{1-a} - \frac{\Delta a}{1+a}\right| = \frac{2\Delta a}{1-a^2}$$ $$\Delta r = r \times \frac{2\Delta a}{1-a^2} = \left(\frac{1-a}{1+a}\right) \times \frac{2\Delta a}{(1-a)(1+a)} = \frac{2\Delta a}{(1+a)^2}$$

Integer and Subjective

33. The side of a cube is measured by vernier callipers (10 divisions of a vernier scale coincide with 9 divisions of main scale, where 1 division of main scale is 1 mm). The main scale reads 10 mm and first division of vernier scale coincides with the main scale. Mass of the cube is 2.736 g. Find the density of the cube in appropriate significant figures. [JEE 2005]

Solution:

$$\text{Least count} = 1\text{ mm} - \frac{9}{10}\text{ mm} = 0.1\text{ mm}$$ $$\text{Side of cube} = 10\text{ mm} + 1 \times 0.1\text{ mm} = 10.1\text{ mm} = 1.01\text{ cm}$$ $$\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{2.736\text{ g}}{(1.01\text{ cm})^3} = \frac{2.736}{1.0303} = 2.655\text{ g/cm}^3 \approx 2.66\text{ g/cm}^3$$

34. During Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2}\text{ m}$ and $3.25 \times 10^{-2}\text{ m}$ of the main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2}\text{ m}$ and $3.25 \times 10^{-2}\text{ m}$ of the main scale but now the 45th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is $8 \times 10^{-7}\text{ m}^2$. The least count of the Vernier scale is $1.0 \times 10^{-5}\text{ m}$. The maximum percentage error in the Young's modulus of the wire is: [JEE Adv. 2014 P-I]

Answer: 4

Solution:

Elongation $\Delta L = (45 - 20) \times \text{L.C.} = 25 \times \text{L.C.}$.

Maximum error in measurement of elongation $\delta(\Delta L) = 1\text{ L.C.}$.

$$\text{Percentage error in } Y = \frac{\delta(\Delta L)}{\Delta L} \times 100 = \frac{1\text{ L.C.}}{25\text{ L.C.}} \times 100 = 4\%$$

35. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density $\rho$ of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to $S^{1/n}$. The value of n is: [JEE Adv. 2014 P-I]

Answer: 3

Solution:

$$d \propto \rho^x S^y f^z$$ $$[\text{L}] = [\text{M}\text{L}^{-3}]^x [\text{M}\text{T}^{-3}]^y [\text{T}^{-1}]^z$$ $$x + y = 0 \implies x = -y$$ $$-3x = 1 \implies x = -\frac{1}{3} \implies y = \frac{1}{3}$$ $$-3y - z = 0 \implies z = -3y = -1$$ $$d \propto S^{1/3} \implies n = 3$$

36. The energy of a system as a function of time t is given as $E(t) = A^2 \exp(-\alpha t)$, where $\alpha = 0.2\text{ s}^{-1}$. The measurement of A has an error of 1.25 %. If the error in the measurement of time is 1.50 %, the percentage error in the value of $E(t)$ at $t = 5\text{ s}$ is: [JEE Adv. 2015 P-II]

Answer: 4

Solution:

$$E = A^2 e^{-\alpha t}$$ $$\frac{\Delta E}{E} = 2 \frac{\Delta A}{A} + \alpha t \left(\frac{\Delta t}{t}\right)$$ $$\%\text{ error in } E = 2(1.25\%) + (0.2 \times 5)(1.50\%) = 2.5\% + 1.5\% = 4\%$$

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