PhysicsQ1–3 of 3 questions
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1
PhysicsMediumMCQAIIMS · 1994
$A$ constant pressure air thermometer gave a reading of $47.5$ units of volume when immersed in ice cold water, and $67$ units in a boiling liquid. The boiling point of the liquid will be .......... $^\circ C$.
A
$135$
B
$125$
C
$112$
D
$100$
Solution
(C) According to Charles's law, for a gas at constant pressure, the volume $V$ is directly proportional to the absolute temperature $T$ ($V \propto T$ or $\frac{V}{T} = \text{constant}$).
Given:
Initial volume $V_1 = 47.5$ units at temperature $T_1 = 0^\circ C = 273 \, K$.
Final volume $V_2 = 67$ units at boiling point temperature $T_2$.
Using the relation $\frac{V_1}{T_1} = \frac{V_2}{T_2}$:
$T_2 = \frac{V_2 \times T_1}{V_1}$
$T_2 = \frac{67 \times 273}{47.5} \approx 385 \, K$.
To convert the temperature from Kelvin to Celsius:
$t(^\circ C) = T(K) - 273$
$t = 385 - 273 = 112^\circ C$.
Thus, the boiling point of the liquid is $112^\circ C$.
Given:
Initial volume $V_1 = 47.5$ units at temperature $T_1 = 0^\circ C = 273 \, K$.
Final volume $V_2 = 67$ units at boiling point temperature $T_2$.
Using the relation $\frac{V_1}{T_1} = \frac{V_2}{T_2}$:
$T_2 = \frac{V_2 \times T_1}{V_1}$
$T_2 = \frac{67 \times 273}{47.5} \approx 385 \, K$.
To convert the temperature from Kelvin to Celsius:
$t(^\circ C) = T(K) - 273$
$t = 385 - 273 = 112^\circ C$.
Thus, the boiling point of the liquid is $112^\circ C$.
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2
PhysicsMediumMCQAIIMS · 1994
$1 \ g$ of ice at $0^\circ C$ is mixed with $1 \ g$ of water at $100^\circ C$. The resulting temperature will be .......... $^\circ C$.
A
$5$
B
$0$
C
$10$
D
$100$
Solution
(C) Let the final temperature be $T$.
Heat lost by $1 \ g$ of water cooling from $100^\circ C$ to $T$ is $Q_{lost} = m_w c_w (100 - T) = 1 \times 1 \times (100 - T) = 100 - T \ \text{cal}$.
Heat gained by $1 \ g$ of ice at $0^\circ C$ to melt and then heat up to $T$ is $Q_{gained} = m_i L_f + m_i c_w (T - 0) = 1 \times 80 + 1 \times 1 \times T = 80 + T \ \text{cal}$.
Equating heat lost and heat gained: $100 - T = 80 + T$.
$2T = 20$,which gives $T = 10^\circ C$.
Heat lost by $1 \ g$ of water cooling from $100^\circ C$ to $T$ is $Q_{lost} = m_w c_w (100 - T) = 1 \times 1 \times (100 - T) = 100 - T \ \text{cal}$.
Heat gained by $1 \ g$ of ice at $0^\circ C$ to melt and then heat up to $T$ is $Q_{gained} = m_i L_f + m_i c_w (T - 0) = 1 \times 80 + 1 \times 1 \times T = 80 + T \ \text{cal}$.
Equating heat lost and heat gained: $100 - T = 80 + T$.
$2T = 20$,which gives $T = 10^\circ C$.
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3
PhysicsEasyMCQAIIMS · 1994
When a body moves with a constant speed along a circle,
A
no work is done on it
B
no acceleration is produced in it
C
its velocity remains constant
D
no force acts on it
Solution
(A) When a body moves with a constant speed along a circle,it undergoes uniform circular motion.
$1$. The force acting on the body is the centripetal force,which is always directed towards the center of the circle.
$2$. The displacement of the body at any instant is along the tangent to the circular path.
$3$. Since the centripetal force is perpendicular to the displacement $(F \perp s)$,the work done $W = F \cdot s \cdot \cos(90^{\circ}) = 0$.
$4$. Therefore,no work is done on the body.
$5$. Acceleration is present (centripetal acceleration),velocity changes due to change in direction,and a centripetal force acts on the body.
$1$. The force acting on the body is the centripetal force,which is always directed towards the center of the circle.
$2$. The displacement of the body at any instant is along the tangent to the circular path.
$3$. Since the centripetal force is perpendicular to the displacement $(F \perp s)$,the work done $W = F \cdot s \cdot \cos(90^{\circ}) = 0$.
$4$. Therefore,no work is done on the body.
$5$. Acceleration is present (centripetal acceleration),velocity changes due to change in direction,and a centripetal force acts on the body.
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