Kinetic Theory Test

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Kinetic Theory Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The average thermal energy for a mono-atomic gas is : ($$K_{B}$$ is Boltzmann constant and $$T$$ is absolute temperature)

A
$$ \dfrac{3}{2} K_{B} T $$

B
$$ \dfrac{5}{2} K_{B} T $$

C
$$ \dfrac{7}{2} K_{B} T $$

D
$$ \dfrac{1}{2} K_{B} T $$

Solution

Average thermal energy $$ = \dfrac{3}{2} K_{B} T $$
where $$3$$ is translational degree of freedom
For monoatomic gas total degree of freedom $$f = 3 $$ (translational degree of freedom)
Question 2
1 / -0

A 0.1kg steel ball falls from a height of 10m and bounces to a height of 7m. The rise in temperature of the ball is:
(Given, $$C=0.1kcal/kg/c^{0}$$)

A
$$0.05^{0}C$$

B
$$0.064^{0}C$$

C
$$0.06^{0}C$$

D
$$0.07^{0}C$$

Solution

Change in potential energy of the ball$$=$$Energy absorbed by the ball

$$mg(h_1-h_2)=mc\Delta T$$

$$\displaystyle \Delta T=\frac {(10)(10-7)}{0.1\times 4.2\times 1000}=0.071^oC$$
Question 3
1 / -0

A bullet travelling at 100 $$ms^{-1}$$ suddenly hits a concrete wall. If its K.E. is converted completely into heat, the raise in temperature is $$\left ( s=100Jkg^{-1}K^{-1} \right )$$ :

A
20K

B
40K

C
50K

D
60K

Solution

Kinetic energy of the bullet is given by:
$$KE = \dfrac{1}{2}mv^{2}$$
The kinetic energy completely changes into heat.
$$Q = ms\Delta T$$
$$KE = Q$$
$$\dfrac{1}{2}mv^{2} = ms\Delta T$$
$$\Delta T = \dfrac{v^{2}}{2s} = \dfrac{100^{2}}{2 \times 100} = 50K$$

Hence option C is correct.
Question 4
1 / -0

2kg of Ice block should be dropped from ‘x km’ height to melt completely. The 8 kg of ice block should be dropped from height

A
4x km

B
x km

C
2x km

D
3x

Solution

Work done by gravity is being converted into heat that is melting the ice ,
$$\therefore$$ we have $$m\times g\times h = m\times L$$
$$\Rightarrow h = L/g $$
Where $$L$$ is latent heat of fusion. From here we can see that height is independent of mass.
So, 8 kg if ice block should be dropped from height $$x$$.
Question 5
1 / -0

From which height a block of ice must be dropped in order that it melts completely. Assume that all the energy is retained by the ice .
($$g=10ms^{-2}$$, $$L=80Jgm^{-1}$$and J $$=$$ 4.2 J/cal)

A
1000Km

B
100Km

C
33.6Km

D
1Km

Solution

The entire potential energy of the waterfall will be converted into heat energy.
$$ mgH = JmL $$
$$ 10 \times H = 80 *4.2 *10^3$$
$$\Rightarrow H = 33.6 Km$$
Question 6
1 / -0

A piece of lead falls from a height of 100m on a fixed non-conducting slab which brings it to rest. The temperature of the lead piece immediately after collision increases by (Sp.heat of lead is 30.6cal/kg/$$^{0}C$$ and $$g=9.8m/sec^{2}.$$)

A
0K

B
$$27^{0}C$$

C
7.62K

D
4.2K

Solution

When the lead ball falls on the non-conducting surface, the lost potential energy is absorbed by the ball as heat energy.

$$mgh=mc\Delta T$$

$$\displaystyle \Delta T=\frac {(9.8)(100)}{(30.6\times 4.2)}=7.625K$$
Question 7
1 / -0

A closed vessel contains some gas at a given temperature and pressure. If the vessel is given a very high velocity, the temperature of the gas

A
increases

B
decreases

C
may increase or decrease depending upon the nature of the gas

D
does not change

Solution

Pressure and Temperature are two thermodynamic parameters which are controlled by microscopic parameters of the system i.e speed of the molecules, their collision with the boundary and other molecules.
At a given temperature molecules move in a random fashion due to thermal energy. so when we move the container in a certain fashion that does mean random motion becomes a systematic motion i.e molecules are still moving in a random fashion and there is no change in motion of molecules with respect to each other and the container, hence no change in Temperature of the Gas.
Question 8
1 / -0

Two metal balls of same material having masses 50gm and 100gm collides with a target with same velocity. Then the ratio of their rise in temperature is

A
1:2

B
4:1

C
2:1

D
1:1

Solution

The rise in the temperature of the target will be proportional to the kinetic energies of the balls.
Since, the are travelling with the same velocity, the rise in temperature
will be proportional to their masses.
Hence, $$ \frac{{T}_{1}}{T_{2}} = \frac{1}{2} $$
Question 9
1 / -0

A bullet of mass $$10\times 10^{-3}\ kg$$ moving with a speed of $$20 ms^{-1}$$ hits an ice block $$(0^{o}c)$$ of $$990\ g$$ kept at rest on a frictionless floor and gets embedded in it. If ice takes $$50\%$$ of $$K.E$$, for melting, then the mass of the ice block that has melted is($$J=4.2 J/Cal$$) (Latent heat of ice $$=80\ cal/g$$)

A
$$6$$

B
$$3$$

C
$$6\times 10^{-3}$$

D
$$3\times 10^{-3}$$

Solution

Kinetic energy of the bullet is $$=\dfrac{1}{2} \times m \times v^2= 2J$$
Now half of it is used by the ice to melt it, i.e. $$1J= 1/4.2$$ cal
Latent heat of the ice is $$L=80cal/gram$$.
Let M is the amount of ice melts:
$$M \times L=1/4.2$$
$$M=0.003 $$gm
And ice will only melt. The temperature will be same.
Question 10
1 / -0

The height of water fall is 210 m assuming that the surface on which the water is falling is perfectly insulated and all the kinetic energy of water is dissipated as heat. Find the rise in temperature of the water :
($$g=10m/s^{2}$$, Specific heat of water $$=1000 \ cal.Kg^{-1}C^{-1}$$ , 1 kcal$$=$$ 4200 J)

A
$$0.1^{0}C$$

B
$$0.5^{0}C$$

C
$$1^{0}C$$

D
$$0.25^{0}C$$

Solution

The entire potential energy of the waterfall will be converted into heat energy.

$$ mgH = mC \delta T $$

$$ 10 \times 210 = 4200 \times \delta T $$

Thus, the rise in temperature will be close to $$0.5^o C$$.

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Kinetic Theory Test - 21

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Kinetic Theory Test - 21

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Question 1

$$ \dfrac{3}{2} K_{B} T $$

$$ \dfrac{5}{2} K_{B} T $$

$$ \dfrac{7}{2} K_{B} T $$

$$ \dfrac{1}{2} K_{B} T $$

Solution

Question 2

$$0.05^{0}C$$

$$0.064^{0}C$$

$$0.06^{0}C$$

$$0.07^{0}C$$

Solution

Question 3

20K

40K

50K

60K

Solution

Question 4

4x km

x km

2x km

3x

Solution

Question 5

1000Km

100Km

33.6Km

1Km

Solution

Question 6

0K

$$27^{0}C$$

7.62K

4.2K

Solution

Question 7

increases

decreases

may increase or decrease depending upon the nature of the gas

does not change

Solution

Question 8

1:2

4:1

2:1

1:1

Solution

Question 9

$$6$$

$$3$$

$$6\times 10^{-3}$$

$$3\times 10^{-3}$$

Solution

Question 10

$$0.1^{0}C$$

$$0.5^{0}C$$

$$1^{0}C$$

$$0.25^{0}C$$

Solution

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