Properties of Matter — Page 10 | NEET Physics

Q91

A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to :

A

{{3F} \over {\left( {\pi {r^2}YT} \right)}}

B

{{6F} \over {\left( {\pi {r^2}YT} \right)}}

C

{F \over {\left( {3\pi {r^2}YT} \right)}}

D

{9F\left( {\pi {r^2}YT} \right)}

Correct Answer

Option A

Solution

Change in length due to temperature change,

\Delta

l

=

l

$\alpha$

\Delta

T

{{\Delta l} \over l}

= $\alpha$ T [ Here

\Delta

T = T ] Y =

{{{F \over {\pi {r^2}}}} \over {{{\Delta l} \over l}}}

=

{{{F \over {\pi {r^2}}}} \over {\alpha T}}

$\Rightarrow$ Y =

{F \over {\pi {r^2}\alpha T}}

$\Rightarrow$ $\alpha$ =

{F \over {\pi {r^2}YT}}

We know, The coefficient of volume expansion ( $\gamma$ ) = 3 $\alpha$ $\therefore$ $\gamma$ =

{{3F} \over {\pi {r^2}YT}}

Q92

A water drop of radius

1 \mathrm{~cm}

is broken into 729 equal droplets. If surface tension of water is 75 dyne/

\mathrm{cm}

, then the gain in surface energy upto first decimal place will be : (Given

\pi=3.14

)

A

8.5 \times 10^{-4} \mathrm{~J}

B

8.2 \times 10^{-4} \mathrm{~J}

C

7.5 \times 10^{-4} \mathrm{~J}

D

5.3 \times 10^{-4} \mathrm{~J}

Correct Answer

Option C

Solution

729 \times {4 \over 3}\pi {r^3} = {4 \over 3}\pi {R^3}

\Rightarrow R = 9r

........ (1)

\Delta U = S \times \Delta A

..... (2)

\Rightarrow \Delta U = S \times \{ - 4\pi {R^2} + 729 \times 4\pi {r^2}\}

= S \times 4\pi \{ 729{r^2} - 81{r^2}\}

= 7.5 \times {10^{ - 4}}\,J

Q93

A steel wire of length

3.2 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{s}}=2.0 \times 10^{11} \,\mathrm{Nm}^{-2}\right)

and a copper wire of length

4.4 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{c}}=1.1 \times 10^{11} \,\mathrm{Nm}^{-2}\right)

, both of radius

1.4 \mathrm{~mm}

are connected end to end. When stretched by a load, the net elongation is found to be

1.4 \mathrm{~mm}

. The load applied, in Newton, will be:

\quad\left(\right.

Given

\pi=\dfrac{22}{7}

)

A 360

B 180

C 1080

D 154

Correct Answer

Option D

Solution

\Delta {l_s} + \Delta {l_c} = 1.4

{{W{l_s}} \over {{Y_s} \times A}} + {{W{l_c}} \over {{Y_c} \times A}} = 1.4 \times {10^{ - 3}}

W = {{1.4 \times {{10}^{ - 3}}} \over {\left[ {{{3.2} \over {2 \times {{(\pi \times 1.4 \times {{10}^{ - 3}})}^2}}} + {{4.4} \over {1.1 \times {{(\pi \times 1.4 \times {{10}^{ - 3}})}^2}}}} \right]{1 \over {{{10}^{ + 11}}}}}}

W \simeq 154\,N

Q94

With rise in temperature, the Young's modulus of elasticity :

A changes erratically

B increases

C decreases

D remains unchanged

Correct Answer

Option C

Solution

The Young's modulus of elasticity, denoted as $E$ , is a measure of the stiffness of a material.

It defines the relationship between stress (force per unit area) and strain (deformation) in a material in the linear (elastic) portion of the stress-strain curve.

As temperature changes, the interatomic distances and bonding energies within a material also change, affecting its mechanical properties, including its elasticity.

For most materials, as the temperature increases, the atoms within the material gain kinetic energy and vibrate more vigorously.

This increased vibration results in a reduction of the forces between atoms, making it easier for the material to deform under a given load.

Hence, generally, the Young's modulus $E$ decreases with increasing temperature because the material becomes softer and less stiff.

Thus, the correct option here would be: Option C: decreases However, it should be noted that the exact relationship between temperature and Young's modulus can vary depending on the material type and its microstructure.

While the general trend for metals and polymers is a decrease in Young's modulus with rising temperature, the rate of decrease and the temperature range over which this occurs can differ significantly between materials.

Some advanced materials and composites may exhibit more complex behavior due to their unique properties.

Q95

A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of

10^5 \mathrm{~N}

at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement

\theta

of the rod axis from its original position would be : (shear moduli,

G=10^{10} \mathrm{~N} / \mathrm{m}^2

)

A

1 / 160 \pi

B

1 / 2 \pi

C

1 / 4 \pi

D

1 / 40 \pi

Correct Answer

Option A

Solution

\begin{aligned} & \text{ Shear moduli }=\frac{\sigma_{\text{shear }}}{\theta} \\ & 10^{10}=\frac{10^5}{\pi \times 16 \times 10^{-4}} \times \frac{1}{\theta} \\ & \theta=\frac{1}{160 \pi} \text{ Radian } \end{aligned}

Q96

A uniformly tapering conical wire is made from a material of Young’s modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3 R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal :

A L

\left( {1 + {2 \over 9}{{Mg} \over {\pi Y{R^2}}}} \right)

B L

\left( {1 + {1 \over 3}{{Mg} \over {\pi Y{R^2}}}} \right)

C L

\left( {1 + {1 \over 9}{{Mg} \over {\pi Y{R^2}}}} \right)

D L

\left( {1 + {2 \over 3}{{Mg} \over {\pi Y{R^2}}}} \right)

Correct Answer

Option B

Solution

Here r = 3R $-$

{{2R} \over L}

x $\therefore$ Extension in the wire of length dx, dl =

{{Fdx} \over {AY}}

=

{{Mg\,dx} \over {\pi {r^2}\,Y}}

=

{{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}

$\therefore$ Change in wire length,

\Delta

L =

\int\limits_0^L {dl}

=

\int\limits_0^L {{{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}}

=

{{Mg} \over {\pi Y}}\int\limits_0^L {{{dx} \over {{{\left( {3R - {{2R} \over L}x} \right)}^2}}}}

=

{{Mg} \over {\pi Y}}\left[ { - {1 \over {\left( {3R - {{2R} \over L}x} \right)}} \times \left( { - {L \over {2R}}} \right)} \right]_0^L

=

{{Mg} \over {\pi Y}}

\left[ {\left( {{L \over {2{R^2}}} - {L \over {6{R^2}}}} \right)} \right]

=

{{Mg} \over {\pi Y}}\left( {{{2L} \over {6{R^2}}}} \right)

=

{{MgL} \over {3\pi {R^2}Y}}

$\therefore$ The equilibrium extended length of the wire, = L +

\Delta

L = L +

{{MgL} \over {3\pi {R^2}Y}}

= L (1 +

{1 \over 3}

{{Mg} \over {\pi {R^2}Y}}

)

Q97

A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take, density of water = 103 kg/m3]

A 30.1 kg

B 87.5 kg

C 65.4 kg

D 46.3 kg

Correct Answer

Option B

Solution

Given

{\left( {50} \right)^3} \times {{30} \over {100}} \times \left( 1 \right) \times g = {M_{cube}}g

...(i) Let m mass should be placed Hence (50)3 $\times$ (1) $\times$ g = (Mcube + m)g …(ii) equation (ii) – equation (i) $\Rightarrow$ mg = (50)3 × g(1 – 0.3) = 125 × 0.7 × 103 g $\Rightarrow$ m = 87.5 kg