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Physics M.C.Q (1 Marks)

Heat and Temperature · UP Board (UPMSP) STD 11 Science (UPMSP - English Medium). 16 questions.

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M.C.Q (1 Marks)

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16 questions · auto-graded multiple-choice test.

MCQ 11 Mark
Consider the following statements.
$A.$ The coefficient of linear expansion has dimension $K^{-1}$.
$B.$ The coefficient of volume expansion has dimension $K^{-1}$.
  • ✓
    $A$ and $B$ are correct.
  • B
    $A$ is correct but $B$ is wrong.
  • C
    $B$ is correct but $A$ is wrong.
  • D
    $A$ and $B$ both wrong.
Answer
Correct option: A.
$A$ and $B$ are correct.
The coefficient of linear expansion,
$\alpha=\frac{1}{\text{L}}\frac{\Delta\text{L}}{\Delta\text{T}}$
$=\frac{|\text{L}|}{|\text{LT}|}=\text{K}^{-1}$
Here, $L =$ initial length
$\Delta\text{L}=$ change in length
$\Delta\text{T}=$ change in temperature
On the other hand, the coefficient of volume expansion,
$\gamma\frac{1}{\text{V}}\frac{\Delta\text{V}}{\Delta\text{T}}=\frac{[\text{L}^3]}{[\text{L}^3\text{T}]}=\text{K}^{-1}$
Here, $V =$ initial volume
$\Delta\text{V}=$ change in volume
$\Delta\text{T}=$ change in temperature
$K =$ kelvin, the $S.I$. unit of temperature.
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MCQ 21 Mark
A metal sheet with a circular hole is heated. The hole:
  • ✓
    Gets larger.
  • B
    Gets smaller.
  • C
    Remains of the same size.
  • D
    Gets deformed.
Answer
Correct option: A.
Gets larger.
When a metal sheet is heated, it starts expanding and its surface area will start increasing, which will lead to an increase in the radius of the hole. Hence, the circular hole will become larger.
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MCQ 31 Mark
Two identical rectangular strips, one of copper and the other of steel, are rivetted together to form a bimetallic strip $(\alpha_{\text{copper}}>\alpha_\text{steel})$. On heating, this strip will:
  • A
    Remain straight.
  • ✓
    Bend with copper on convex side.
  • C
    Bend with steel on convex side.
  • D
    Get twisted.
Answer
Correct option: B.
Bend with copper on convex side.
We are provided with two metal strips of copper and steel. On heating, both of them will expand. Expansion coefficient of copper is more than that of steel. So, the copper metal strip will expand more, causing the bimetallic strip to bend with copper at the convex side, as it'll have more surface area compared to the steel sheet, which will be on the concave side.
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MCQ 41 Mark
A system $X$ is neither in thermal equilibrium with $Y$ nor with $Z$. The systems $Y$ and $Z$:
  • A
    Must be in thermal equilibrium.
  • B
    Cannot be in thermal equilibrium.
  • ✓
    May be in thermal equilibrium.
  • D
    None of these.
Answer
Correct option: C.
May be in thermal equilibrium.
The given data in the question is insufficient to specify the relation between the physical conditions of systems $Y$ and $Z$. As system $X$ is not in thermal equilibrium with $Y$ and $Z$, systems $Y$ and $Z$ may be at the same temperature or they may or may not be in thermal equilibrium with each other.
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MCQ 51 Mark
The temperature of water at the surface of a deep lake is $2^\circ C.$ The temperature expected at the bottom is:
  • A
    $0^\circ C$
  • B
    $2^\circ C$
  • ✓
    $4^\circ C$
  • D
    $6^\circ C$
Answer
Correct option: C.
$4^\circ C$
The density of water is maximum at $4^\circ C,$ and the water at the bottom of the lake is most dense, compared to the layers of water above. Therefore, the temperature expected at the bottom is $4^\circ C.$
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MCQ 61 Mark
If the temperature of a uniform rod is slightly increased by $\Delta\text{t}$, its moment of inertia I about a perpendicular bisector increases by:
  • A
    $\text{Zero}$
  • B
    $\alpha\text{I}\Delta\text{t}$
  • ✓
    $2\alpha\text{I}\Delta\text{t}$
  • D
    $3\alpha\text{I}\Delta\text{t}$
Answer
Correct option: C.
$2\alpha\text{I}\Delta\text{t}$
The change in moment of inertia of uniform rod with change is temperature is given by,
$\text{I}'=\text{I}(1+2\alpha\Delta\text{t})$
Here, I = initial moment of inertia
I' = new moment of inertia due to change in temperature
$\alpha=\text{expansion coefficient}$
$\Delta\text{t}=\text{change in temperature}$
$\text{so},\text{I}'-\text{I}=2\alpha\text{I}\Delta\text{t}$
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MCQ 71 Mark
An aluminium sphere is dipped into water at $10^\circ C.$ If the temperature is increased, the force of buoyancy:​​​​​​
  • A
    Will increase.
  • ✓
    Will decrease.
  • C
    Will remain constant.
  • D
    May increase or decrease depending on the radius of the sphere.
Answer
Correct option: B.
Will decrease.
When an aluminium sphere is dipped in water and the temperature of water is increased, the aluminium will start expanding leading to increase in its volume. This will lead to increase in the surface area of the shell and it'll exert less pressure on the water such that the volume of the sphere submerged in water will decrease and it'll start float easily on water. Now, the volume of water displaced will be less compared to what was displaced initially. Therefore, the force of buoyancy will decrease, as it is directly proportional to the volume of water displaced.
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MCQ 81 Mark
For a constant volume gas thermometer, one should fill the gas at:
  • A
    Low temperature and low pressure.
  • B
    Low temperature and high pressure.
  • ✓
    High temperature and low pressure.
  • D
    High temperature and high pressure.
Answer
Correct option: C.
High temperature and low pressure.
A constant-volume gas thermometer should be filled with an ideal gas in which particles don't interact with each other and are free to move anywhere, so that the thermometer functions properly. An ideal gas is only a theoretical possibility. Therefore, the gas that is filled in the thermometer should be at high temperature and low pressure, as under these conditions, a gas behaves as an ideal gas.
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MCQ 91 Mark
As the temperature is increased, the time period of a pendulum.
  • A
    Increases proportionately with temperature.
  • ✓
    Increases.
  • C
    Decreases.
  • D
    Remains constant.
Answer
Correct option: B.
Increases.
In general, the time period of a pendulum, $t,$ is given by
$\text{t}=\frac{1}{2\pi}\sqrt\frac{\text{l}}{\text{g}}$
When the temperature $(T)$ is increased, the length of the pendulum $(l)$ is given by,
$\text{l}=\text{l}_0(1+\alpha\text{T}),$
where $l_0 =$ length at $0^\circ C$
$\alpha=$ linear coefficient of expansion
Therefore, the time period of a pendulum will be
$\text{t}=\frac{1}{2\pi}\sqrt\frac{\text{l}_0(1+\alpha\text{T})}{\text{g}}$
Hence, time period of a pendulum will increase with increase in temperature.
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MCQ 101 Mark
A spinning wheel $A$ is brought in contact with another wheel $B,$ initially at rest. Because of the friction at contact, the second wheel also starts spinning. Which of the following energies of the wheel $B$ increases?
  • A
    Kinetic.
  • B
    Total and  Internal.
  • C
    Mechanical.
  • ✓
    All of the above
Answer
Correct option: D.
All of the above
When the wheel $B$ starts spinning because of the friction at contact, it will gain kinetic energy and, hence, mechanical energy $($kinetic $+$ potential energies$)$. Also, internal energy will increase, which increases with rise in temperature.
Along with it, the generation of heat energy due to friction will lead to increase in the net sum of all the energies, i.e. total energy.
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MCQ 111 Mark
In which of the following pairs of temperature scales, the size of a degree is identical?
  1. Mercury scale and ideal gas scale.
  2. Celsius scale and mercury scale.
  3. Celsius scale and ideal gas scale.
  4. Ideal gas scale and absolute scale.
  • A
    $A$ and $B$
  • B
    $B$ and $C$
  • C
    Only $C$
  • ✓
    $C$ and $D$
Answer
Correct option: D.
$C$ and $D$
Celsius scale and ideal gas scale measure temperature in kelvin $(K)$ and the ideal gas scale is sometimes also called the absolute scale. A mercury scale gives reading in degrees and its size of degree, which depends on length of mercury column, doesn't match any of the above$-$mentioned scales.
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MCQ 121 Mark
If the temperature of a uniform rod is slightly increased by ∆t, its moment of inertia Iabout a line parallel to itself will increase by:
  • A
    $\text{Zero}$
  • B
    $\alpha\text{I}\Delta\text{t}$
  • ✓
    $2\alpha\text{I}\Delta\text{t}$
  • D
    $3\alpha\text{I}\Delta\text{t}$
Answer
Correct option: C.
$2\alpha\text{I}\Delta\text{t}$
The moment of inertia of a solid body of any shape changes with temperature as
$\text{I}'=\text{I}(1+2\alpha\Delta\text{t})$
Here, I = initial moment of inertia
I' = new moment of inertia due to change in temperature
$\alpha=\text{expansion coefficient}$
$\Delta\text{t}=\text{change in temperature}$
$\text{So},\text{I}'-\text{I}=2\alpha\text{I}\Delta\text{t}$
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MCQ 131 Mark
A spinning wheel is brought in contact with an identical wheel spinning at identical speed. The wheels slow down under the action of friction. Which of the following energies of the first wheel decreases?
  • ✓
    Kinetic and Mechanical.
  • B
    Total.
  • C
    External
  • D
    Internal.
Answer
Correct option: A.
Kinetic and Mechanical.
The kinetic energy of a body depends on its speed. Since when a spinning wheel is slowed down, its speed decreases leading to reduction in its kinetic energy. The mechanical energy of a body is defined as the sum of its potential and kinetic energies. Since the kinetic energy of the wheel has been decreased, it'll lead to decrease in its mechanical energy. When the wheel slows down due to friction, its mechanical energy gets converted into heat energy, leading to increase in internal energy, which increases with increase in temperature.
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MCQ 141 Mark
A body $A$ is placed on a railway platform and an identical body $B$ in a moving train. Which of the following energies of $B$ are greater than those of $A$, as seen from the ground?
  • A
    Kinetic.
  • B
    Total.
  • C
    Mechanical.
  • ✓
    All of the above
Answer
Correct option: D.
All of the above
As body $A$ is at rest on the ground, it possesses only potential energy, whereas body $B,$ being placed inside a moving train, possesses kinetic energy due to its motion along with the train. Therefore, body $B$ will have greater kinetic, mechanical $($energy possessed by the body by virtue of its position and motion $=$ kinetic energy $+$ potential $+$ energy$)$ energy and, hence, total $($sum of all the energies$)$ energy. No information is given about the temperature of the body so we can not say wheather body $\text{B}^\prime$ s internal energy will be or will not be greater than that of body $A.$
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MCQ 151 Mark
A solid object is placed in water contained in an adiabatic container for some time. The temperature of water falls during this period and there is no appreciable change in the shape of the object. The temperature of the solid object:
  • ✓
    Must have increased.
  • B
    Must have decreased.
  • C
    May have increased.
  • D
    May have remained constant.
Answer
Correct option: A.
Must have increased.
The whole system (water + solid object) is enclosed in an adiabatic container from which no heat can escape. After some time, the temperature of water falls, which implies that the heat from the water has been transferred to the object, leading to increase in its temperature.
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MCQ 161 Mark
Which of the following pairs may give equal numerical values of the temperature of a body?
  • ✓
    Fahrenheit and Kelvin.
  • B
    Celsius and Kelvin.
  • C
    Kelvin and Platinum.
  • D
    None of these
Answer
Correct option: A.
Fahrenheit and Kelvin.
We know that the relation between the temperature in Fahrenheit and Kelvin scales is given by
$\frac{\text{T}_\text{F}-32}{180}=\frac{\text{T}_\text{K}-273.15}{100}$
$\text{T}_\text{F}=\text{T}_\text{K}=\theta$
Therefore,
$\frac{\theta-32}{180}=\frac{\theta-273.15}{100}$
$5\theta-160=9\theta-2758.5$
$4\theta=2298.35$
$\theta=574.59^\circ$
If we consider the same for Celsius and Kelvin scales
$\frac{\text{T}_\text{C}-0}{100}=\frac{\text{T}_\text{K}-273.15}{100}$
Let the temperature be t
$\frac{\text{t}-0}{100}=\frac{\text{t}-273.15}{100}$
$\text{t}=\text{t}-273.15$
Thus, $t$ does not exist.
The Kelvin scale uses mercury as thermometric substance, whereas the platinum scale uses platinum as thermometric substance. The scale depends on the properties of the thermometric substance used to define the scale. The platinum and Kelvin scales do not agree with each other. Therefore, there is no such temperature that has same numerical value in the platinum and Kelvin scale.
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