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Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

The units of fore and length are made three times of their earlier values. Earlier the energy of a system was $81\ J$. What will be the energy of the same system in new units?

  1. $243$

  2. $729$

  3. $9$

  4. $None\ of\ the\ above$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy has dimensions [M L^2 T^-2]. If force [M L T^-2] and length [L] are tripled, then [M L T^-2] -> 3 and [L] -> 3. This implies [M] -> 3 / 3 = 1 and [T^-2] -> 3 / 3 = 1 (no change). Thus, energy scales by [M] * [L]^2 * [T^-2] = 1 * 3^2 * 1 = 9. New energy = 81 * 9 = 729 J.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

$1$ calorie is the heat required to increase the temperature of $1g$ of water by $1 ^ { \circ } \mathrm { C }$ from:

  1. $13.5 ^ { \circ } \mathrm { C } \text { to } 14.5 ^ { \circ } \mathrm { C } \text { at } 76 \mathrm { mm } \text { of } \mathrm { Hg }$

  2. $14.5 ^ { \circ } \mathrm { C } \text { to } 15.5 ^ { \circ } \mathrm { C } \text { at } 760 \mathrm { mm } \text { of } \mathrm { Hg }$

  3. $13.5 ^ { \circ } \mathrm { C } \text { to } 15.5 ^ { \circ } \mathrm { Cat } 76 \mathrm { mm } \text { of } \mathrm { Hg }$

  4. $15.5 ^ { \circ } \mathrm { C } \text { to } 16.5 ^ { \circ } \mathrm { C } \text { at } 700 \mathrm { mm } \text { of } \mathrm { Hg }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

If the amount of heat given to a system is $35\, J$ and the amount of work done on the system is $15\, J$, then the change in internal energy of the system is

  1. $- 50\, J$

  2. $20\, J$

  3. $30\, J$

  4. $50\, J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

$\Delta Q=+35J$
$\Delta W=-15J$
$\Delta U=?$
From law of thermodynamic,
$\Delta Q=\Delta U+\Delta W$
$\Delta U=\Delta Q-\Delta W$
$\Delta U=35-(-15)$
$\Delta U=35+15$
$\Delta U=50J$
The correct option is D. 

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A geyser heats water flowing at the rate of 3.0 liters per minute from ${ 27 }^{ \circ  }C$ to ${ 77 }^{ \circ  }C$. If the geyser operates on a gas burner, the rate of consumption of the fuel if its heat of combustion is $4.0\times { 10 }^{ 4 }J/g$ per minute is

  1. $15.75g$

  2. $4 g$

  3. $0.3 g$

  4. $0.16 g$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Heat required = mass * specific heat * delta T = 3000 g * 1 cal/g C * 50 C = 150,000 cal = 630,000 J. Fuel consumption rate = Total heat / Heat of combustion = 630,000 J / 40,000 J/g = 15.75 g.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A certain quantity of heat energy is given to a diatomic ideal gas which expands at constant pressure. The fraction of the heat energy that is converted into work is 

  1. $\dfrac 2 5$

  2. $\dfrac 2 7$

  3. $\dfrac 1 5$

  4. $\dfrac 5 7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Supplied heat at constant pressure, 

$\Delta Q=nC _P \Delta T$
Change in internal energy,
$\Delta U=nC _V \Delta T$
ratio, $\dfrac{\Delta U}{\Delta Q}=\dfrac{nC _V \Delta T}{nC _P \Delta T}=\dfrac{C _V}{C _P}$. . . . . . . . .(1)
For diatomic ideal gas,
$C _P=\dfrac{7R}{2}\,  ,  C _V=\dfrac{5R}{2}$
From equation (1), we get
$\dfrac{\Delta U}{\Delta Q}=\dfrac{5R/2}{7R/2}=\dfrac{5}{7}$
$(5/7)^{th}$ part of heat supplied is used to increase internal energy.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

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. $12.1 ^ { \circ } \mathrm { C }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial kinetic energy = 0.5 * m * v^2 = 0.5 * 1 * 50^2 = 1250 J. Half of this is 625 J. Convert to calories: 625 / 4.2 = 148.8 cal. Heat = ms(delta T) -> 148.8 = 200 g * 0.105 cal/g C * delta T. delta T = 148.8 / 21 = 7.08 C.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

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$\mathrm { kJ }$
Reveal answer Fill a bubble to check yourself
A Correct answer