Units & Measurements — Page 4 | JEE Physics

Q31

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?

A 0.70 mm

B 0.50 mm

C 0.75 mm

D 0.80 mm

Correct Answer

Option D

Solution

Least count =

{{0.5} \over {50}}

= 0.01 mm Zero error = (45 - 50) $\times$ 0.01 mm = - 0.05 mm Thickness of sheet = (0.5 + 25 $\times$ 0.01) - (-0.05) = 0.50 + 0.30 = 0.80 mm

Q32

In a Young's double slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is :

A 3

B 6

C 12

D 24

Correct Answer

Option C

Solution

In a double slit experiment, the bright spots (maxima) from interference are found where: $d\,\sin \theta = m\lambda$ , where $m$ can be 0, ±1, ±2, and so on.

The dark spots (minima) from single slit diffraction appear at: $a\,\sin \theta = \pm\lambda$ .

For a bright interference fringe to be inside the central bright area of the single slit (central maximum), it must fall between these two first diffraction minima.

This means we need: $|m\lambda/d| \leq \lambda/a$ , or $|m| \leq d/a$ .

The problem says the distance between the slits ( $d$ ) is 6.1 times the width of each slit ( $a$ ), so $d/a = 6.1$ .

The biggest whole number for $m$ is 6, so there are maxima for $m = \pm1, \pm2, ..., \pm6$ .

If we count only the bright spots on the sides (not the middle one), that's 6 on one side and 6 on the other, making 12 bright maxima.

If the central bright spot ( $m=0$ ) is counted too, there would be 13, but usually “within the central maximum” means only the fringes between the two minima, not the central line.

So, you will see 12 bright interference fringes inside the central maximum of the single slit envelope.

Answer = 12.

(Option C)

Q33

A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If half life of C14 is 5730 years, then age of the wooden piece placed in the museum is approximately :

A 10439 years

B 13094 years

C 19039 years

D 39049 years

Correct Answer

Option C

Solution

Fresh wood: 20 decays/min (current activity if alive / just cut).

Old museum sample: 2 decays/min.

Half-life of C‑14 = 5730 years.

Find the age of wooden piece in museum.

Step 1: Ratio of activities $\dfrac{A}{A_0} = \dfrac{2}{20} = 0.1$ So, the museum sample’s activity is 10% of that in fresh wood.

Step 2: Radioactive decay law $A = A_0 e^{-\lambda t}$ $\dfrac{A}{A_0} = e^{-\lambda t}$ So, $0.1 = e^{-\lambda t}$ $t = \dfrac{\ln(0.1)}{-\lambda}$ Step 3: Decay constant $\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{5730 \,\text{yr}} \approx 1.2097 \times 10^{-4} \,\text{yr}^{-1}$ Step 4: Solve for $t$ $t = \dfrac{\ln(0.1)}{-1.2097 \times 10^{-4}}$ $\ln(0.1) = -2.3026$ $t = \dfrac{2.3026}{1.2097 \times 10^{-4}}$ $t \approx 19039 \,\text{years}$ ✅ Final Answer: Option C: 19039 years

Q34

The diameter of the objective lens of microscope makes an angle

\beta

at the focus of the microscope. Further, the medium between the object and the lens is an oil of refractive index n. Then the resolving power of the microscope.

A Increases with decreasing value of n

B Increases with decreasing value of

\beta

C Increases with increasing value of n sin 2

\beta

D Increases with increasing value of

{1 \over {n\sin 2\beta }}

Correct Answer

Option C

Solution

The smallest detail a microscope can see (its resolution limit) is given by the formula:

d_{\text{min}} \approx \frac{\lambda}{2 \text{N.A.}}

Here, N.A. stands for numerical aperture. The formula for numerical aperture is:

\text{N.A.} = n \sin \alpha

where n is the refractive index of the oil (the medium), and α is half the angle made by the cone of light rays coming from the specimen.

The resolving power (RP) of the microscope tells us how well it can see small details.

It is the inverse of the resolution limit:

\text{RP} = \frac{1}{d_{\text{min}}} \propto n \sin \alpha

Now, the angle subtended by the lens diameter at the focal point is β.

From geometry, we see that α (half the cone angle) is related to β by:

\sin \alpha = \sin 2\beta

This means the resolving power becomes:

\text{RP} \propto n \sin 2\beta

So, the microscope will see finer details as the value of $n \sin 2\beta$ increases. Answer: C.

Q35

A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is

\theta

iC and the Brewster's angle of incidence is

\theta

iB, such that sin

\theta

iC/sin

\theta

iB =

\eta

= 1.28. The relative refractive index of the two media is :

A 0.2

B 0.4

C 0.8

D 0.9

Correct Answer

Option C

Solution

We are tasked with finding the relative refractive index given the relation between the critical angle and the Brewster’s angle.

Step 1: Recall the definitions Critical angle (from denser medium $n_1$ to rarer medium $n_2$ , with $n_1 > n_2$ ): $\sin \theta_c = \dfrac{n_2}{n_1} = \dfrac{1}{\mu}$ where $\mu = \dfrac{n_1}{n_2}$ is the relative refractive index $> 1$ .

Brewster’s angle: $\tan \theta_B = \dfrac{n_2}{n_1} = \dfrac{1}{\mu}$ Step 2: Given condition $\dfrac{\sin \theta_c}{\sin \theta_B} = \eta = 1.28$ So: $\dfrac{1/\mu}{\sin \theta_B} = 1.28$ Step 3: Relation between $\theta_B$ and $\mu$ Since $\tan \theta_B = \tfrac{1}{\mu}$ , $\sin \theta_B = \dfrac{\tan \theta_B}{\sqrt{1+\tan^2 \theta_B}} = \dfrac{\tfrac{1}{\mu}}{\sqrt{1+\tfrac{1}{\mu^2}}} = \dfrac{1}{\sqrt{\mu^2+1}}$ Step 4: Substitute into ratio $\dfrac{\sin \theta_c}{\sin \theta_B} = \dfrac{\tfrac{1}{\mu}}{\tfrac{1}{\sqrt{\mu^2+1}}} = \dfrac{\sqrt{\mu^2+1}}{\mu}$ Given this equals 1.28: $\dfrac{\sqrt{\mu^2+1}}{\mu} = 1.28$ Step 5: Solve for $\mu$ $\sqrt{\mu^2+1} = 1.28 \mu$ Square both sides: $\mu^2 + 1 = 1.6384 \, \mu^2$ $1 = (1.6384 - 1)\mu^2$ $1 = 0.6384 \mu^2$ $\mu^2 = \dfrac{1}{0.6384} \approx 1.565$ $\mu \approx 1.25$ Step 6: Find the relative index of the two media in the question’s terms Here we defined $\mu = \dfrac{n_1}{n_2} \approx 1.25$ .

But the relative refractive index of the two media usually means $\dfrac{n_2}{n_1}$ (rarer/denser), since that's how it appears in options (<1).

$\dfrac{n_2}{n_1} = \dfrac{1}{\mu} \approx \dfrac{1}{1.25} = 0.8$ ✅ Final Answer: $\boxed{0.8}$ Correct Option: C

Q36

The force is given in terms of time t and displacement x by the equation F = A cos Bx + C sin Dt The dimensional formula of

{{AD} \over B}

is :

A

[{M^0}L{T^{ - 1}}]

B

[M{L^2}{T^{ - 3}}]

C

[{M^1}{L^1}{T^{ - 2}}]

D

[{M^2}{L^2}{T^{ - 3}}]

Correct Answer

Option B

Solution

[A] = [ML{T^{ - 2}}]

[B] = [{L^{ - 1}}]

[D] = [{T^{ - 1}}]

\left[ {{{AD} \over B}} \right] = {{[ML{T^{ - 2}}][{T^{ - 1}}]} \over {[{L^{ - 1}}]}}

\left[ {{{AD} \over B}} \right] = [M{L^2}{T^{ - 3}}]

Q37

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

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 and main scale has 10 divisions in 1 cm.

Correct Answer

Option D

Solution

Measured length of rod = 3.50 cm That means least count of the measuring instrument should be 0.01 cm = 0.1 mm For vernier scale 1 main scale division = 1 mm And 9 MSD = 10 VSD Least count = 1 MSD - 1 VSD = 1 - 0.9 = 0.1 mm

Q38

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 1 s, then the reported mean time should be :

A 92

\pm

1.8 s

B 92

\pm

3 s

C 92

\pm

2 s

D 92

\pm

5.0 s

Correct Answer

Option C

Solution

Here t1 = 90 s, t2 = 91 s, t3 = 95 s, t4 = 92 s Mean(t) =

{{{t_1} + {t_2} + {t_3} + {t_4}} \over 4}

=

{{90 + 91 + 95 + 92} \over 4}

= 92 s Now mean deviation =

{{2 + 1 + 3 + 0} \over 4}

= 1.5 s Since least count of clock is one second, so reported mean time = (92 $\pm$ 2) s

Q39

A physical quantity P is described by the relation P = a

^{{{1}/{2}}}

b2 c3 d

-

4 If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be :

A 8%

B 12%

C 32%

D 25%

Correct Answer

Option C

Solution

Given, P = a

^{{1 \over 2}}

b2 c3 d $-$ 4 Relative error =

{{\Delta P} \over P}

$\times$ 100 = (

{1 \over 2}

$\times$

{{\Delta a} \over a}

+ 2

{{\Delta b} \over b}

+ 3

{{\Delta c} \over c}

+ 4

{{\Delta d} \over d}

) $\times$ 100 =

{1 \over 2}

$\times$ 2 + 2 $\times$ 1 + 3 $\times$ 3 + 4 $\times$ 5 = 32%

Q40

The following observations were taken for determining surface tension T of water by capillary method: diameter of capillary, D = 1.25

\times

10-2 m rise of water, h = 1.45

\times

10-2m Using g = 9.80 m/s2 and the simplified relation T =

{{rhg} \over 2} \times {10^3}N/m

, the possible error in surface tension is closest to :

A 10 %

B 0.15 %

C 1.5 %

D 2.4 %

Correct Answer

Option C

Solution

Surface tension, T =

{{rhg} \over 2} \times {10^3}N/m

Relative error,

{{\Delta T} \over T} = {{\Delta r} \over r} + {{\Delta h} \over h}

Percentage error,

{{\Delta T} \over T} \times 100 = {{\Delta r} \over r} \times 100 + {{\Delta h} \over h} \times 100

{{\Delta T} \over T} \times 100 = \left( {{{{{10}^{ - 2}} \times 0.01} \over {1.25 \times {{10}^{ - 2}}}} + {{{{10}^{ - 2}} \times 0.01} \over {1.45 \times {{10}^{ - 2}}}}} \right) \times 100

= (0.8 + 0.689) = 1.489 % = 1.5 %