Tag: the essence of change

Questions Related to the essence of change

A hammer of mass 1$\mathrm { kg }$ having speed of 50$\mathrm { m } / \mathrm { s }$ , hit a iron nail of mass 200$\mathrm { gm }$ . If specific heat of iron is 0.105 cal/gm'C and half the energy is converted into heat, the raise in temperature of nail is

  1. $7.1 ^ { \circ } C$

  2. $9.2 ^ { \circ } \mathrm { C }$

  3. $10.5 ^ { \circ } C$

  4. $12.1 ^ { \circ } \mathrm { C }$


Correct Option: A

In case of explosion of a bomb which of the following does not change?

  1. Chemical energy

  2. Total energy

  3. Kinetic energy

  4. Mechanical energy


Correct Option: B
Explanation:

The chemical energy possessed by the bomb after explosion converts into heat energy, sound energy, light energy i.e. into different forms of energy but remains conserved. Hence, total energy of bomb does not change, but kinetic and mechanical energy changes.

In the foregoing question, the amount of energy given to the system from A to B is:

  1. 50J

  2. 60J

  3. 30J

  4. 10J


Correct Option: A

The specific heat capacity of a metal at low temperautre (T) is given as $C _ { p } \left( k \sqrt { k } - 1 k g ^ { - 1 } \right) = 32 \left( \frac { T } { 400 } \right) ^ { 3 }$ A 100 gram vessel of this metal is to be cooled from $20 ^ { \circ } K$ to $4 ^ { \circ } \mathrm { K }$ by a special refrigerator operating at room temperature $\left( 27 ^ { \circ } \mathrm { C } \right) .$ The amount of work required to cool the vessel is:-

  1. equal to 0.002$\mathrm { kJ }$

  2. greater than 0.148$\mathrm { kJ }$

  3. between 0.148$\mathrm { kJ }$ and 0.028$\mathrm { kJ }$

  4. less than 0.028$\mathrm { kJ }$


Correct Option: A

An ideal gas heat engine operates in a Carnot's cycle between $227^{\circ}C$ and $127^{\circ}C$. It absorbs $6\times 10^{4} J$ at high temperature. The amount of heat converted into work is _____.

  1. $4.8\times 10^{4}J$

  2. $3.5\times 10^{4}J$

  3. $1.6\times 10^{4}J$

  4. $1.2\times 10^{4}J$


Correct Option: D

Work done by 100 calorie of heat is __________.

  1. 418.4 J

  2. 4.184 J

  3. 41.84 J

  4. None of these


Correct Option: A
Explanation:
For an isothermal expansion of an ideal gas, the change in internal energy is zero.

According to the first law of thermodynamics, 

Change in internal energy U = Q-W = 0

So, all the heat energy is utilized to do work. 

Q = W

We know that, one calorie is equal to 4.184 J

Therefore, Work done by 100 calorie of heat in an isothermal expansion of any ideal gas will be 4.184 * 100 =  418.4 Joule


A person of weight 60 kg wants to loose 5 kg by going up and down 10m high stairs. Assume he burns twice as much fat while going up than going down. If 1 kg of fat is burnt on expending 7000 kcal. How many times must he go up and down to reduce his 7 kg weight? (Take $  g=10 \mathrm{ms}^{-2} )  $

  1. $ 1.8 \times 10^{3} $ times

  2. $ 2.4 \times 10^{3} $ times

  3. $ 1.7 \times 10^{3} $ times

  4. $ 2.1 \times 10^{3} $ times


Correct Option: C
Explanation:
Energy used to go up $=mgh=60\times 10\times 10=6000\,J$

Energy used to come down $\dfrac{6000}{2}=3000\,J$

Energy used in one round trip $=9000\,J$

$1\,cal=4.5\,J$

$1\,J=\dfrac{1}{4.2\,cal}$

$9000\,J=\dfrac{9000}{4.2}=2142.85\,cal$

$7000\,kilo\,cal$ is required to burn $1\,kg$ mass

To reduce $5\,kg$ mass, energy required $=7000\times 5=35000\,kilo\,val$

Number of trip $=\dfrac{35000\times 1000}{2142.85}=1.7\times 10^{3}$

What is $\Delta E$  for a system that does 5 kcal of work by the surroundings when 3 kcal heat is absorbed by the system.  

  1. -2 kcal

  2. +2 kcal

  3. +8kcal

    • 7 kcal

Correct Option: C

A bullet of mass 10 gm moving with a speed of 20 m/s hits an ice block of mass 990 gm kept on a frictionless floor and gets stuck in it. The amount of ice that melts, if 50% of the lost kinetic energy goes to ice, will be 

  1. $0.003 g$

  2. $0.30 g$

  3. $0.0003 g$

  4. $3.0 g$


Correct Option: A

A steel ball of mass $5$ ${ g }$ is thrown downward with velocity $10$ ${ ms } ^ { - 1 }$ from height $19.5$ ${ m }$ . It penetrates sand by $50$ ${ cm }$ . The change in mechanical energy will be ( ${ g } = 10$ ${ ms } ^ { - 2 }$ )

  1. $1$ ${J}$

  2. $1.25$ ${J}$

  3. $1.5$ ${J}$

  4. $1.75$ ${J}$


Correct Option: B
Explanation:

$\begin{array}{l} The\, \, change\, \, in\, \, mechanic\, \, energy\, \, \Delta U=mg\left( { h+x } \right) +\frac { 1 }{ 2 } m{ v^{ 2 } } \ here\, \, m=5g=0.00\, 5kg\cdot h=19.5\, mx=50cm=0.5m,v=10\, m/s \ So,\, \Delta U=0.005\times 10919.5+0.5+\frac { 1 }{ 2 } \times 0.005\times { \left( { 10 } \right) ^{ 2 } }=0.005\times 10\times 20+\frac { 1 }{ 2 } \times 0.005\times 100=1.25J \end{array}$

Hence,
option $(B)$ is correct answer.