Properties of Matter — Page 9 | NEET Physics

Q81

A black body has maximum wavelength

\lambda

m at 2000 K. Its corresponding wavelength at 3000 K will be

A

{3 \over 2}{\lambda _m}

B

{2 \over 3}{\lambda _m}

C

{{16} \over {81}}{\lambda _m}

D

{{81} \over {16}}{\lambda _m}

Correct Answer

Option B

Solution

According to Wein's displacement law,

{\lambda _m}T

= 2.88 × 10 –3 When T = 2000 K,

{\lambda _m}

(2000) = 2.88 × 10 –3 ....(1) When T = 3000 K,

\lambda _m^{'}

(3000) = 2.88 × 10 –3 ....(2) Dividing (1) by (2),

{2 \over 3}{{{\lambda _m}} \over {\lambda _m^{'}}} = 1

\Rightarrow {{{\lambda _m}} \over {\lambda _m^{'}}} = {3 \over 2}

\Rightarrow \lambda _m^{'} = {2 \over 3}{\lambda _m}

Q82

A solid steel ball of diameter 3.6 mm acquired terminal velocity

2.45 \times 10^{-2} \mathrm{~m} / \mathrm{s}

while falling under gravity through an oil of density

925 \mathrm{~kg} \mathrm{~m}^{-3}

. Take density of steel as

7825 \mathrm{~kg} \mathrm{~m}^{-3}

and g as

9.8 \mathrm{~m} / \mathrm{s}^2

. The viscosity of the oil in SI unit is

A 2.18

B 1.68

C 2.38

D 1.99

Correct Answer

Option D

Solution

To determine the viscosity of the oil, we use the formula for the terminal velocity of a sphere falling through a viscous fluid: $v_T = \dfrac{2}{9} \dfrac{(\rho_s - \rho_f) r^2 g}{\eta}$ Where: $v_T$ is the terminal velocity, $\rho_s$ is the density of the steel ball, $\rho_f$ is the density of the fluid, $r$ is the radius of the ball, $g$ is the acceleration due to gravity, $\eta$ is the viscosity of the fluid.

Given: Diameter of the steel ball = 3.6 mm, therefore radius $r = 1.8$ mm = $1.8 \times 10^{-3}$ m, Terminal velocity $v_T = 2.45 \times 10^{-2}$ m/s, Density of oil $\rho_f = 925$ kg/m $^3$ , Density of steel $\rho_s = 7825$ kg/m $^3$ , Acceleration due to gravity $g = 9.8$ m/s $^2$ .

Rearranging the formula to solve for $\eta$ : $\eta = \dfrac{2}{9} \cdot \dfrac{(\rho_s - \rho_f) \cdot r^2 \cdot g}{v_T}$ Substituting the known values into the equation: $\eta = \dfrac{2}{9} \cdot \dfrac{(7825 - 925) \cdot (1.8 \times 10^{-3})^2 \cdot 9.8}{2.45 \times 10^{-2}}$ Calculating this, we approximate the viscosity $\eta$ to be: $\eta \approx 1.99 \, \text{Pa}\cdot\text{s}$ Thus, the viscosity of the oil is approximately 1.99 Pa·s.

Q83

Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Clothes containing oil or grease stains cannot be cleaned by water wash. Reason (R): Because the angle of contact between the oil/grease and water is obtuse. In the light of the above statements, choose the correct answer from the option given below.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Correct Answer

Option A

Solution

Due to obtuse angle of contact the water doesn't wet the oiled surface properly and cannot wash it also.

$\Rightarrow$ Assertion is correct and Reason given is a correct explanation.

Q84

Under the same load, wire A having length

5.0 \mathrm{~m}

and cross section

2.5 \times 10^{-5} \mathrm{~m}^{2}

stretches uniformly by the same amount as another wire B of length

6.0 \mathrm{~m}

and a cross section of

3.0 \times 10^{-5}

\mathrm{m}^{2}

stretches. The ratio of the Young's modulus of wire A to that of wire

B

will be :

A

1: 2

B

1: 4

C

1: 1

D

1: 10

Correct Answer

Option C

Solution

$\Delta \ell=\dfrac{F \ell}{S Y}$ $F$ is same for both wire and $\Delta \ell$ is also same

\begin{aligned} & \frac{\Delta \ell}{F}=\frac{\ell}{S Y} \\\\ & \Rightarrow \frac{\ell_{A}}{S_{A} Y_{A}}=\frac{\ell_{B}}{S_{B} Y_{B}} \\\\ & \Rightarrow \frac{5}{2.5 \times Y_{A}}=\frac{6}{3 \times Y_{B}} \\\\ & \Rightarrow \frac{Y_{A}}{Y_{B}}=1 \end{aligned}

Q85

A small spherical ball of radius

r

, falling through a viscous medium of negligible density has terminal velocity '

v

'. Another ball of the same mass but of radius

2 r

, falling through the same viscous medium will have terminal velocity:

A

4 \mathrm{v}

B

2 \mathrm{~V}

C

\dfrac{v}{4}

D

\dfrac{\mathrm{v}}{2}

Correct Answer

Option D

Solution

Since density is negligible hence Buoyancy force will be negligible At terminal velocity.

\mathrm{Mg} =6 \pi \eta \mathrm{rv}

\mathrm{V} \propto \frac{1}{\mathrm{r}} \quad

(as mass is constant) Now,

\frac{\mathrm{v}}{\mathrm{v}^{\prime}}=\frac{\mathrm{r}^{\prime}}{\mathrm{r}}

r^{\prime}=2 \mathrm{r}

So,

v^{\prime}=\frac{v}{2}

Q86

A capillary tube of radius 0.1 mm is partly dipped in water (surface tension 70 dyn/cm and glass water contact angle ≈ 0°) with 30° inclined with the vertical. The length of water risen in the capillary is _______ cm. (Take

g = 9.8 \text{ m/s}^2

)

A

\dfrac{82}{5}

B

\dfrac{68}{5}

C

\dfrac{57}{2}

D

\dfrac{71}{5}

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}=\frac{2 \times 70 \times 1}{1 \times 980 \times 10^{-2}} \\ & \mathrm{~h}=\frac{100}{7} \mathrm{~cm} \\ & \sin 60^{\circ}=\frac{\mathrm{h}}{\ell} \\ & \ell=\frac{\mathrm{h} \times 2}{\sqrt{3}} \\ & \ell=\frac{100}{7} \times \frac{2}{\sqrt{3}} \\ & =\frac{200}{7 \times \sqrt{3}} \\ & =16.49 \mathrm{~cm} \end{aligned}

Q87

In an experiment to verify Stokes law, a small spherical ball of radius r and density

\rho

falls under gravity through a distance h in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of h is proportional to : (ignore viscosity of air)

A r

B r4

C r3

D r2

Correct Answer

Option B

Solution

After falling through h, the velocity be equal to terminal velocity

\sqrt {2gh}

=

{2 \over 9}{{{r^2}g} \over \eta }\left( {{\rho _l} - \rho } \right)

$\Rightarrow$ h =

{2 \over {81}}{{{r^4}g{{\left( {{\rho _l} - \rho } \right)}^2}} \over {{\eta ^2}}}

$\Rightarrow$ h $\propto$ r4

Q88

A

20

cm

long capillary tube is dipped in water. The water rises up to

8

cm.

If the entire arrangement is put in a freely falling elevator the length of water column in the capillary tube will be

A

10

cm

B

8

cm

C

20

cm

D

4

cm

Correct Answer

Option C

Solution

In freely falling elevator

g

= 0 Water fills the tube entirely in gravity less condition.

Hence, length of water column in the capillary tube is 20 cm.

Q89

A body takes 10 minutes to cool from 60oC to 50oC. The tempertature of surroundings is constant at 25oC. Then, the temperature of the body after next 10 minutes will be approximately :

A 47oC

B 41oC

C 45oC

D 43oC

Correct Answer

Option D

Solution

Time taken to cool from 60

^\circ

C to 50

^\circ

C = 10 minutes Temperature of surroundings = 25

^\circ

C Temperature of body in next 10 minutes = T Therefore,

{{60 - 50} \over {10\,\min }} = {k_B}\left[ {{{60 + 50} \over 2} - 25} \right] \Rightarrow {k_B}30 = 1

...... (1) and

{{60 - T} \over {20\,\min }} = {k_B}\left[ {{{60 + T} \over 2} - 25} \right] = {k_B}\left[ {{{60 + T - 50} \over 2}} \right]

..... (2) Taking ratio of Eqs. (1) and (2), we get

{{20} \over {60 - T}} = {{30{k_B}} \over {{k_B}\left( {{{10 + T} \over 2}} \right)}} \Rightarrow {{20} \over {60 - T}} = {{30} \over {5 + T/2}}

\Rightarrow 20\left( {5 + {T \over 2}} \right) = 30(60 - T) \Rightarrow 100 + T10 = 1800 - 30T

\Rightarrow 1800 - 100 = 30T + 10T

\Rightarrow 1700 = 40T \Rightarrow T = {{1700} \over {40}} = 42.5^\circ C \sim 43^\circ C

Q90

A submarine experiences a pressure of 5.05 × 106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 × 106 Pa. Then d2 –d1 is approximately (density of water = 103 kg/m3 and acceleration due to gravity = 10 ms–2 ) :

A 600 m

B 400 m

C 300 m

D 500 m

Correct Answer

Option C

Solution

The pressure experienced by a submarine at a certain depth in the sea is given by the formula: $P = \rho g h$ where: $P$ is the pressure $\rho$ is the density of the fluid (sea water in this case) $g$ is the acceleration due to gravity $h$ is the height (or depth in this case) Given: $\rho = 10^3 \, kg/m^3$ $g = 10 \, m/s^2$ We are looking for the difference in depth, $d_2 - d_1$ , which corresponds to the difference in pressure $\Delta P$ : $\Delta P = P_2 - P_1 = \rho g (d_2 - d_1)$ Rearranging the above equation, we get: $d_2 - d_1 = \dfrac{\Delta P}{\rho g}$ Given: $P_2 = 8.08 \times 10^6 \, Pa$ $P_1 = 5.05 \times 10^6 \, Pa$ $\Delta P = P_2 - P_1 = 8.08 \times 10^6 \, Pa - 5.05 \times 10^6 \, Pa = 3.03 \times 10^6 \, Pa$ So, $d_2 - d_1 = \dfrac{3.03 \times 10^6 \, Pa}{10^3 \, kg/m^3 \times 10 \, m/s^2} = 303 \, m$ The closest answer among the options provided is Option C, 300 m.