Properties of Matter — Page 5 | JEE Physics

Q41

A thin 1 m long rod has a radius of 5 mm. A force of 50

\pi

kN is applied at one end to determine its Young’s modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false ?

A

{{\Delta \gamma } \over \gamma }

gets minimum contribution from the uncertainty in the length.

B The figure of merit is the largest for the length of the rod.

C The maximum value of

\gamma

that can be determined is 2

\times

1014 N/m2

D

{{\Delta \gamma } \over \gamma }

gets its maximum contribution from the uncertainty in strain

Correct Answer

Option C

Solution

Young's Modulus of the material of the rod is

Y = {{Stress} \over {Strain}} = {{(F/A)} \over {(\Delta l/l)}}

Here, Y remains maximum, when

\Delta

l is of least count. That is,

{Y_{\max }} = \left[ {{{50\pi \times {{10}^3}N} \over {\pi {{(5 \times {{10}^{ - 3}})}^2}{m^2}}}} \right]\left[ {{{1m} \over {0.01 \times {{10}^{ - 3}}m}}} \right]

= 2 \times {10^4}N/{m^2}

Q42

Two tubes of radii r1 and r2, and lengths l1 and l2 , respectively, are connected in series and a liquid flows through each of them in stream line conditions. P1 and P2 are pressure differences across the two tubes. If P2 is 4P1 and l2 is

{{{1_1}} \over 4}

, then the radius r2 will be equal to :

A r1

B 2r1

C 4r1

D

{{{r_1}} \over 2}

Correct Answer

Option D

Solution

We know, Rate of flow of liquid through narrow tube,

{{dv} \over {dt}}

=

{{\pi {{\Pr }^4}} \over {8\eta l}}

Both tubes are connected in series so rate of flow of liquid is same.

\therefore\,\,\,

{{\pi {P_1}r_1^4} \over {8\eta {l_1}}}

=

{{\pi {P_2}r_2^4} \over {8\eta {l_2}}}

$\Rightarrow$

\,\,\,

{{{P_1}{r_1}^4} \over {{l_1}}}

=

{{{P_2}{r_2}^4} \over {{l_2}}}

Given, P2 = 4P1 and

{l_2}

=

{{{l_1}} \over 4}

$\Rightarrow$

\,\,\,

{{{P_1}{r_1}^4} \over {{l_1}}}

=

{{4{P_1}{r_2}^4} \over {{{{l_1}} \over 4}}}

$\Rightarrow$

\,\,\,

{r_2}^4

=

{{{r_1}^4} \over {16}}

$\Rightarrow$

\,\,\,

r2 =

{{{r_1}} \over 2}

Q43

A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of :

A

{1 \over {81}}

B 9

C

{1 \over {9}}

D 81

Correct Answer

Option B

Solution

To determine how the stress in the leg changes when a man's linear dimensions increase by a factor of 9, we can analyze the relationship between the dimensions and the stress on the legs.

First, let's establish some relationships: Linear dimensions (length, width, height) increase by a factor of 9.

Density remains the same.

Stress is defined as force per unit area:

\text{Stress} = \dfrac{\text{Force}}{\text{Area}}

Since density remains constant, the volume and thus the mass of the man will change according to the cube of the linear dimensions.

Since the linear dimension changes by a factor of 9, the volume (and hence the mass) changes by a factor of:

9^3 = 729

Therefore, the weight (force due to gravity) also increases by a factor of 729.

The cross-sectional area of the leg is proportional to the square of the linear dimensions.

Thus, if the linear dimension increases by a factor of 9, the cross-sectional area increases by a factor of:

9^2 = 81

Now, substituting these factors into the stress formula:

\text{New Stress} = \dfrac{\text{New Force}}{\text{New Area}} = \dfrac{729 \times \text{Original Force}}{81 \times \text{Original Area}}

Simplifying, we get:

\text{New Stress} = 9 \times \dfrac{\text{Original Force}}{\text{Original Area}} = 9 \times \text{Original Stress}

Thus, the stress in the leg increases by a factor of 9. Therefore, the correct answer is: Option B: 9

Q44

A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let P2 be the pressure inside the inner bubble and P0, the pressure outside the outer bubble. Radius of another bubble with pressure difference P2

-

P0 between its inside and outside would be :

A 12 cm

B 2.4 cm

C 6 cm

D 4.8 cm

Correct Answer

Option B

Solution

Pressure difference inside the inner bubble, p2 $-$ p1 =

{{4T} \over {{r_2}}}

b . . . . . (1) And for outer bubble p1 $-$ p0 =

{{4T} \over {{r_1}}}

. . . . . . . (2)

\therefore\,\,\,

p2 $-$ p0 = 4T

\left( {{1 \over {{r_2}}} + {1 \over {{r_1}}}} \right)

$\Rightarrow$

\,\,\,

p2 $-$ p0 =

{{4T} \over r}

Here r is the radius of the bubble.

\therefore\,\,\,

{1 \over r} = {1 \over {{r_2}}} + {1 \over {{r_1}}}

$\Rightarrow$

\,\,\,

r =

{{{r_1}{r_2}} \over {{r_1} + {r_2}}}

=

{{4 \times 6} \over {4 + 60}}

= 2.4 cm

Q45

Young's moduli of two wires A and B are in the ratio 7 : 4. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to :-

A 1.7 mm

B 1.9 mm

C 1.3 mm

D 1.5 mm

Correct Answer

Option A

Solution

Given:

{{{Y_A}} \over {{Y_B}}} = {7 \over 4};\,{L_A} = 2m\,;{A_A} = \pi {R^2}

{F \over A} = Y\left( {{l \over L}} \right);{L_B} = 1.5m\,;{A_B} = \pi {(2mm)^2}

given F and

l

are same $\Rightarrow$

{{AY} \over L}

is same

{{{A_A}{Y_A}} \over {{L_A}}} = {{{A_B}{Y_B}} \over {{L_B}}}

\Rightarrow {{\left( {\pi {R^2}} \right)\left( {{7 \over 4}{Y_B}} \right)} \over 2} = {{\pi {{(2\,mm)}^2}.{Y_B}} \over {1.5}}

R = 1.74 mm

Q46

A thin uniform tube is bent into a circle of radius

r

in the vertical plane. Equal volumes of two immiscible liquids, whose densities are

{\rho _1}

and

{\rho _2}

\left( {{\rho _1} > {\rho _2}} \right),

fill half the circle. The angle

\theta

between the radius vector passing through the common interface and the vertical is :

A

\theta = {\tan ^{ - 1}}\pi \left( {{{{\rho _1}} \over {{\rho _2}}}} \right)

B

\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1}} \over {{\rho _2}}}} \right)

C

\theta = {\tan ^{ - 1}}\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)

D

\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right)

Correct Answer

Option C

Solution

As system is in equilibrium so the pressuse at A from both side of the liquid must be equal . (r cos $\theta$ + r sin $\theta$ ) $\rho$ 2g = (r cos $\theta$ $-$ r sin $\theta$ ) $\rho$ 1g $\Rightarrow$

\,\,\,

{{{\rho _1}} \over {{\rho _2}}} = {{\sin \theta + \cos \theta } \over {\cos \theta - \sin \theta }} = {{1 + \tan \theta } \over {1 - \tan \theta }}

$\Rightarrow$

\,\,\,

$\rho$ 1 $-$ $\rho$ 1 tan $\theta$ = $\rho$ 2 + $\rho$ 2 tan $\theta$ $\Rightarrow$

\,\,\,

( $\rho$ 1 + $\rho$ 2) tan $\theta$ = $\rho$ 1 $-$ $\rho$ 2 $\Rightarrow$

\,\,\,

tan $\theta$ =

{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}

$\Rightarrow$

\,\,\,

$\theta$ = tan $-$ 1

\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)

Q47

A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere,

\left( {{dr \over r}} \right)

is:

A

{{mg} \over {Ka}}

B

{{Ka} \over {mg}}

C

{{Ka} \over {3mg}}

D

{{mg} \over {3Ka}}

Correct Answer

Option D

Solution

Because of m mass the extra pressure created is,

\Delta

P =

{{mg} \over a}

And Bulk modulus, $\beta$ =

{{\Delta P} \over {{{\Delta V} \over V}}}

Given $\beta$ = K

\therefore\,\,\,

K =

{{{{mg} \over a}} \over {{{\Delta V} \over V}}}

We know volume of sphere, V =

{4 \over 3}\pi {r^3}

\therefore\,\,\,

{{dV} \over V}

= 3

{{dr} \over r}

\therefore\,\,\,

K =

{{{{mg} \over a}} \over {3{{dr} \over r}}}

$\Rightarrow$

\,\,\,

{{dr} \over r}

=

{{mg} \over {3Ka}}

Q48

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

Q49

If 'M' is the mass of water that rises in a capillary tube of radius 'r', then mass of water which will rise in a capillary tube of radius '2r' is :

A M

B 4M

C M/2

D 2M

Correct Answer

Option D

Solution

Height of liquid rise in capillary tube

h = {{2T\,\cos {\theta _c}} \over {\rho rg}}

\Rightarrow h \propto {1 \over r}

When radius becomes double height become half $\therefore$

{h^{'}} = {h \over 2}

Now, M = $\pi$ r2h × $\rho$ and M' = $\pi$ (2r)2 (h/2) × $\rho$ = 2M

Q50

The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of thier densities is 13.6. Their contact angles, with glass, are close to 135° and 0°, respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius r1, while water rises by the same amount h in a capillary tube of radius r2. The ratio, (r1/r2), is then close to :

A 2/5

B 2/3

C 3/5

D 4/5

Correct Answer

Option A

Solution

h = {{2{S_1}\cos \theta } \over {{r_1}{\rho _1}g}}

h = {{2{s_2}\cos {\theta _2}} \over {{r_2}{\rho _2}g}}

\Rightarrow {{{r_1}} \over {{r_2}}} = {2 \over 5}