Properties of Matter — Page 3 | NEET Physics

Q21

The unit of thermal conductivity is :

A J m –1 K –1

B W m K –1

C J m K –1

D W m –1 K –1

Correct Answer

Option D

Solution

We know, Heat transfer (Q) = KA

\left( {{{\Delta T} \over {\Delta x}}} \right)

$\times$ t Here, K is the coefficient of thermal conductivity J = (K)m 2

{{kelvin} \over m}

$\times$ s $\therefore$ Unit of K = Wm –1 K –1

Q22

When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L +

l

). The elastic potential energy stored in the extended wire is :

A Mgl

B mgL

C

{1 \over 2}

Mgl

D

{1 \over 2}

MgL

Correct Answer

Option C

Solution

Elastic potential energy U =

{1 \over 2}

(work done due to gravity) =

{1 \over 2}

Mgl

Q23

A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5

\times

10 –2 N/m. The pressure inside the bubble equals at a point Z 0 below the free surface of water in a container. Taking g = 10 m/s 2 , density of water = 10 3 kg/m 3 , the value of Z 0 is :

A 1 cm

B 10 cm

C 100 cm

D 0.5 cm

Correct Answer

Option A

Solution

P = P 0 + $\rho$ gZ 0 .........(i) Also, P = P 0 +

{{4T} \over R}

.....(ii) From (i) and (ii), $\rho$ gZ 0 =

{{4T} \over R}

$\therefore$ Z 0 =

{{4T} \over {\rho gR}}

=

{{4 \times 2.5 \times {{10}^{ - 2}}} \over {{{10}^3} \times 10 \times {{10}^{ - 3}}}}

= 10 -2 m = 1 cm

Q24

A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in tmperature. The length of aluminium rod is : (

\alpha

Cu = 1.7 × 10 –5 K –1 and

\alpha

Al = 2.2 × 10 –5 K –1 )

A 119.9 cm

B 88 cm

C 68 cm

D 6.8 cm

Correct Answer

Option C

Solution

At any temperature

{\left( {\Delta l} \right)_{Cu}} = {\left( {\Delta l} \right)_{Al}}

{l_{Cu}}{\alpha _{Cu}}\Delta T = {l_{Al}}{\alpha _{Al}}\Delta T

$\Rightarrow$ 88 $\times$ 1.7 $\times$ 10 -5 =

{l_{Al}}

$\times$ 2.2 × 10 –5 $\Rightarrow$

{l_{Al}}

= 68 cm

Q25

A small hole of area of cross-section 2 mm 2 present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s 2 , the rate of flow of water through the open hole would be nearly :

A 8.9 × 10 –6 m 3 /s

B 2.23 × 10 –6 m 3 /s

C 6.4 × 10 –6 m 3 /s

D 12.6 × 10 –6 m 3 /s

Correct Answer

Option D

Solution

volume flow rate = Av = A

\sqrt {2gh}

= 2 $\times$ 10 -6 $\times$

\sqrt {2 \times 10 \times 2}

= 2 $\times$ 2 $\times$ 3.14 $\times$ 10 -6 m 3 /s = 12.56 $\times$ 10 -6 m 3 /s = 12.6 $\times$ 10 -6 m 3 /s

Q26

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by

\Delta

l on applying a force F, how much force is needed to stretch the second wire by the same amount?

A 6F

B F

C 9F

D 4F

Correct Answer

Option C

Solution

Let the wire 1 is of length = l and wire 2 of length =

{l \over 3}

Now area of wire 1 = A while area of wire 2 = 3A For wire 1,

\Delta

l =

\left( {{F \over {AY}}} \right)l

....(i) For wire 2, let F' force is applied

\Delta

l =

\left( {{{F'} \over {3AY}}} \right){l \over 3}

....(2) From equation (i) & (ii)

\left( {{F \over {AY}}} \right)l

=

\left( {{{F'} \over {3AY}}} \right){l \over 3}

$\Rightarrow$ F' = 9F

Q27

The power radiated by a black body is P and it radiates maximum energy at wavelength,

\lambda

0 . If the temperature of the black body is now changed so that it radiates maximum energy at wavelength

{3 \over 4}{\lambda _0}

, the power radiated by it becomes nP. The value of n is

A

{3 \over 4}

B

{4 \over 3}

C

{{256} \over {81}}

D

{{81} \over {256}}

Correct Answer

Option C

Solution

From Wien's law, $\lambda$ max T = constant $\therefore$ $\lambda$ max 1 T 1 = $\lambda$ max 2 T 2 $\Rightarrow$ $\lambda$ 0 T 1 =

{{3{\lambda _0}} \over 4}

T 2 $\Rightarrow$

{{T_2} \over T_1} = {4 \over 3}

According to Stefan-Boltzmann law, energy emitted unit time by a black body is Ae $\sigma$ T 4 , $\therefore$ P $\propto$ T 4 So

{{{P_2}} \over {{P_1}}} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4}

=

{\left( {{4 \over 3}} \right)^4}

Here, P 2 = nP and P 1 = P So,

{{nP} \over P} = n = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4} = {\left( {{4 \over 3}} \right)^4} = {{256} \over {81}}

Q28

A small sphere of radius ‘r’ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

A r 3

B r 2

C r 5

D r 4

Correct Answer

Option C

Solution

Power = rate of production of heat = F.V = 6 $\pi$ $\eta$ rV T .

V T = = 6 $\pi$ $\eta$ rV T 2 [ As F = 6 $\pi$ $\eta$ rV T from stoke’s formula ] $\because$ Terminal velocity V T =

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

$\therefore$ V T $\propto$ r 2 $\therefore$ Power $\propto$ 6 $\pi$ $\eta$ r

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

$\Rightarrow$ Power $\propto$ r 5

Q29

Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are K 1 and K 2 . The thermal conductivity of the composite will be

A

{{3\left( {{K_1} + {K_2}} \right)} \over 2}

B

{K_1} + {K_2}

C

2\left( {{K_1} + {K_2}} \right)

D

{{{K_1} + {K_2}} \over 2}

Correct Answer

Option D

Solution

Equivalent thermal conductivity of the composite rod in parallel combination will be,

K = {{{K_1}{A_1} + {K_2}{A_2}} \over {{A_1} + {A_2}}} = {{{K_1} + {K_2}} \over 2}

Q30

The bulk modulus of a spherical object is 'B'. If it is subjected to uniform pressure 'P', the fractional decrease in radius is

A

{B \over {3p}}

B

{{3p} \over B}

C

{p \over {3B}}

D

{p \over B}

Correct Answer

Option C

Solution

Bulk modulus is given by

B = {P \over {\left( {{{\Delta V} \over V}} \right)}} \Rightarrow {{\Delta V} \over V} = {P \over B}

3{{\Delta R} \over R} = {P \over B}

(here,

{{\Delta R} \over R}

= fractional decreases in radius)

\Rightarrow {{\Delta R} \over R} = {P \over {3B}}