Free Objective Test 02 Practice Test - 11th and 12th

Question 1

Latent heat of ice is 80 calorie/gm. A man melts 60 g of ice by chewing in 1 minute. His power is

A. 4800 W

B. 336 W

C. 1.33 W

D. 0.75 W

SOLUTION

Solution : B

Work done by man = Heat absorbed by ice = mL = 60 $\times$ 80 = 4800 calorie = 20160 J
$∴ P o w e r = \frac{W}{t} = \frac{20160}{60} = 336 W$

Question 2

The graph AB shown in figure is a plot of temperature of a body in degree Celsius and degree Fahrenheit. Then

A. Slope of line AB is

\frac{9}{5}

B. Slope of line AB is

\frac{5}{9}

C. Slope of line AB is

\frac{1}{9}

D. Slope of line AB is

\frac{3}{9}

SOLUTION

Solution : B

Relation between Celsius and Fahrenheit scale of temperature is $\frac{C}{5} = \frac{F - 32}{9}$
By rearranging we get, C = $\frac{5}{9} F - \frac{160}{9}$
By equating above equation with standard equation of line $y = m x + c$ we get $m = \frac{5}{9}$ and $c = \frac{- 160}{9}$
i.e. Slope of the line AB is $\frac{5}{9}$ .

Question 3

Given that P, V and T stand for vapour pressure, volume and temperature, an ideal gas obeys the law -PV α T. If the volume is kept constant, P α T. With decreasing temperature the vapour pressure drops, until P = 0 at some temperature T₀. Since P cannot be less than zero, or negative, T₀naturally becomes the lowest temperature that we can reach, and can be used as a "natural” lower fixed point while constructing a temperature scale. This is the "absolute zero”, which in the Celsius scale is - 273.15⁰C.

To construct an absolute temperature scale , where the absolute zero is naturally 0. Let us choose the triple point of mercury (-38.8⁰C, at 0.2 mPa) as our upper fixed point, and give assign it a value 100. What will be the boiling point of water on this scale?

121.5

110

159.2

182.7

SOLUTION

Solution : C

Let the boiling point of water in the near scale be T. Let's list what we know -

We can, therefore, write -

Question 4

We understand 'temperature' as a measure of the 'hotness' or 'coldness' of a body. Which of these properties do you think explicitly depend on the temperature?

Density of a liquid

length of a metal rod

pressure exerted by a gas inside a closed chamber

all of the above

SOLUTION

Solution : D

Let us look at some real life examples.
(1)Density of water is seen to be 1 kg/m³ only at 4⁰C. At 80⁰C, it is seen to get reduced to 0.971 kg/m³.
(2)A plastic bottle half-filled with water often shrinks when kept inside the refrigerator.
(3)If you keep heating a car tyre, which is made of a certain sturdy rubber, it will explode at some point.
Although it is unscientific to make general statements based on singular observations, we are inclined, at this point, to make the following guesses -
(1)Density of water decreases with increasing temperature
(2)Decreasing temperature shrinks a plastic bottle
(3)Since we can assume the tyre is sturdy and will not leak easily, an explosion will only be due to an increase in the pressure of air inside.

Thus, all the properties listed from (a) to (c) seem to explicitly depend on temperature.

Question 5

A bimetallic strip is formed out of two identical strips, one of copper and other of brass. The coefficients of linear expansion of the two metals are $α_{C}$ and $α_{B}$ . On heating, the temperature of the strip goes up by $Δ$ T and the strip bends to form an arc of radius of curvature R. Then R is

A. Proportional to

Δ

B. Inversely proportional to

Δ

C. Proportional to

| α_{B} - α_{C} |

D. Inversely proportional to

| α_{B} - α_{C} |

SOLUTION

Solution : B and D

On heating, the strip undergoes linear expansion
So after expansion length of brass strip $L_{B} = L_{0} (1 + α_{B} Δ T)$ and length of copper strip $L_{C} = L_{0} (1 + α_{C} Δ T)$

From the figure $L_{B} = (R + d) θ \dots \dots (i)$
and $L_{c} = R θ \dots \dots (i i)$ [As angle = Arc/Radius]
Dividing (i) by (ii) $\frac{R + d}{R} = \frac{L_{B}}{L_{C}} = \frac{1 + α_{B} Δ T}{1 + α_{C} Δ T}$
$\Rightarrow$ $1 + \frac{d}{R} = (1 + α_{B} Δ T) (1 + α_{C} Δ T)^{- 1}$ = $(1 + α_{B} Δ T) (1 - α_{C} Δ T)$ = $1 + (α_{B} - α_{C}) Δ T$
$\Rightarrow$ $\frac{d}{R} = (α_{B} - α_{C}) Δ T$ or $R = \frac{d}{(α_{B} - α_{C}) Δ T}$ [Using Binomial theorem and neglecting higher terms]
So we can say R $\propto \frac{1}{(α_{B} - α_{C})}$ and R $\propto \frac{1}{Δ T}$

Question 6

50 g of copper is heated to increase its temperature by $10^{\circ} C .$ If the same quantity of heat is given to 10 g of water, the rise in its temperature is:
Specific heat of copper is $420 J k g^{- 1}^{\circ} C$ .

5^{\circ} C

6^{\circ} C

7^{\circ} C

8^{\circ} C

SOLUTION

Solution : A

Same amount of heat is supplied to copper and water.
$m_{c} c_{c} Δ T_{c} = m_{w} c_{w} Δ T_{w}$
$Δ T_{w} = \frac{m_{c} c_{c} Δ T_{c}}{m_{w} c_{w}}$
$Δ T_{w} = \frac{50 \times 10^{- 3} \times 420 \times 10}{10 \times 10^{- 3} \times 4200}$
$Δ T_{w} = 5^{\circ} C$

Question 7

A solid whose volume does not change with temperature floats in a liquid. For two different temperatures $t_{1}$ and $t_{2}$ of the liquid, fractions $f_{1}$ and $f_{2}$ of the volume of the solid remain submerged in the liquid. The coefficient of volume expansion of the liquid is equal to

\frac{f_{1} - f_{2}}{f_{2} t_{1} - f_{1} t_{2}}

\frac{f_{1} - f_{2}}{f_{1} t_{1} - f_{2} t_{2}}

\frac{f_{1} + f_{2}}{f_{2} t_{1} + f_{1} t_{2}}

\frac{f_{1} + f_{2}}{f_{1} t_{1} + f_{2} t_{2}}

SOLUTION

Solution : A

As with the rise in temperature, the liquid undergoes volume expansion therefore the fraction of solid submerged in liquid increases.
Fraction of solid submerged at ${t_{1}}^{\circ} C = f_{1}$ = Volume of displaced liquid $= V_{0} (1 + γ t_{1}) \dots \dots (i)$ and fraction of solid submerged at ${t_{2}}^{\circ} C = f_{2}$ = Volume of displaced liquid $= V_{0} (1 + γ t_{2}) \dots \dots (i i)$
From (i) and (ii) $\frac{f_{1}}{f_{2}} = \frac{1 + γ t_{1}}{1 + γ t_{2}}$ $\Rightarrow$ $γ = \frac{f_{1} - f_{2}}{f_{2} t_{1} - f_{1} t_{2}}$

Question 8

A wire of length $L_{0}$ is supplied heat to to raise its temperature by T. If $γ$ is the coefficient of volume ex the wire and Y is the young's modulus of the wire then the energy density stored in the wire is

\frac{1}{2} γ^{2} T^{2} Y

\frac{1}{3} γ^{2} T^{2} Y^{3}

\frac{1}{18} \frac{γ^{2} T^{2}}{Y}

\frac{1}{18} γ^{2} T^{2} Y

SOLUTION

Solution : D

Due to heating the length of the wire increases. $∴$ Longitudinal strain is produced $\Rightarrow \frac{Δ L}{L} = a \times Δ T$
Elastic potential energy per unit volume $E = \frac{1}{2} \times S t r e s s \times S t r a i n = \frac{1}{2} \times Y \times (S t r a i n)^{2}$
$\Rightarrow E = \frac{1}{2} \times Y \times {(\frac{Δ L}{L})}^{2} = \frac{1}{2} \times Y \times a^{2} \times Δ T^{2} o r E = \frac{1}{2} \times Y \times {(\frac{γ}{3})}^{2} \times T^{2} = \frac{1}{18} γ^{2} Y T^{2} [A s γ = 3 a a n d Δ T = T (g i v e n)]$

Question 9

A thermometer is graduated in mm. It registers - 3mm when the bulb of thermometer is in pure melting ice and 22mm when the thermometer is in steam at a pressure of one atm. The temperature in $^{\circ} C$ when the thermometer registers 13mm is

\frac{13}{25} \times 100

\frac{16}{25} \times 100

\frac{13}{22} \times 100

\frac{16}{22} \times 100

SOLUTION

Solution : B

For a constant volume gas thermometer temperature in $^{\circ} c e n t i g r a d e$ is given as
$T_{c} = \frac{P - P_{0}}{P_{100} - P_{0}} \times 100^{\circ} C$ $\Rightarrow$ $T_{c} = \frac{13 - (- 3)}{22 - (- 3)} \times 100^{\circ} C = \frac{16}{25} \times 100$

Question 10

The freezing point on a thermometer is marked as $20^{\circ}$ and the boiling point at as $150^{\circ}$ . A temperature of $60^{\circ}$ C on this thermometer will be read as

40^{\circ}

65^{\circ}

98^{\circ}

110^{\circ}

SOLUTION

Solution : C

Temperature on any scale can be converted into other scale by $\frac{X - L F P}{U F P - L F P}$ = Constant for all scales
$∴$ $\frac{X - 20^{\circ}}{150^{\circ} - 20^{\circ}} = \frac{C - 0^{\circ}}{100^{\circ} - 0^{o}}$ $\Rightarrow$ X = $\frac{C \times 130^{\circ}}{100^{\circ}} + 20^{\circ}$ = $\frac{60^{\circ} \times 130^{\circ}}{100^{\circ}} + 20^{\circ} = 98^{\circ}$

Question 11

The density of a substance at $20^{\circ} C$ and $120^{\circ} C$ are $1.1 g c . c^{- 1}$ and $1 g c . c^{- 1}$ respecttively. Find the coefficient of cubical expansion of the solid.

${0.001}^{\circ} C^{- 1}$

${0.002}^{\circ} C^{- 1}$

{0.003}^{\circ} C^{- 1}

${0.004}^{\circ} C^{- 1}$

SOLUTION

Solution : A

Given: Initial density, $d_{1} = 1.1 g c . c^{- 1}$
Final density, $d_{2} = 1 g c . c^{- 1}$
Initial tempereature, $t_{1} = 20^{\circ} C$
Final temperature, $t_{2} = 120^{\circ} C$

Volumetric expansion of solids can be expressed as,
$V_{2} = V_{1} (1 + γ (t_{2} - t_{1}))$
where, $V_{1}$ is the initial volume, $V_{2}$ is the final volume, $t_{1}$ is the initial temperature, $t_{2}$ is the final temperature and $γ$ is the coefficient of cubical expansion of solid.
We know $v o l u m e = \frac{m a s s}{d e n s i t y}$
Let $d_{1}$ be the density at volume $V_{1}$ and $d_{2}$ be the density at volume $V_{2}$
Since mass of the substance, $m$ remains same, we have,
or, $\frac{m}{d_{2}} = \frac{m}{d_{1}} (1 + γ (t_{2} - t_{1}))$
or, $γ = \frac{d_{1} - d_{2}}{d_{2} (t_{2} - t_{1})}$
or, $γ = \frac{1.1 - 1}{1 (120 - 20)} = \frac{0.1}{100} = 0.001$
$∴ γ = {0.001}^{\circ} C^{- 1}$

Question 12

Which of the following curves represents the relation between the Celsius and Fahrenheit scales?

SOLUTION

Solution : D

The relation between the two scales is given as -

$F = (\frac{9}{5}) C + 32 \Rightarrow C = \frac{5 F}{9} - \frac{(32 \times 5)}{9} .$

The slope of the line is $\frac{5}{9}$ .

When the Fahrenheit scale reads zero, the reading of the Celsius scale is

$C = \frac{5 \times 0}{9} - \frac{(32 \times 5)}{9} = - {17.77}^{\circ} C$

$∴$ The slope is positive, and the line intercepts the negative y-axis; hence, (d) is correct.

Question 13

In a constant volume gas thermometer, the pressure of the working gas is measured by the difference in the levels of mercury in the two arms of a U-tube connected to the gas at one end. When the bulb is placed at the room temperature 27.0℃, the mercury column in the arm open to atmosphere stands 5.00 cms above the level of mercury in the other arm. When the bulb is placed in a hot liquid, the difference of mercury levels becomes 45.0 cms. Calculate the temperature of the liquid. (Atmospheric pressure = 75.0 cm of mercury.)

155℃

177℃

187℃

165℃.

SOLUTION

Solution : B

The pressure of the gas = atmospheric pressure + the pressure due to the difference in mercury levels

At 27℃, the pressure is (75 cm + 5 cm) = 80 cm of mercury. At the liquid's temperature, the pressure is (75 cm + 45 cm) = 120 cm of mercury

Using $T_{2} = (\frac{P_{2}}{P_{1}}) T_{1},$ the temperature of the liquid is -

$T = [(\frac{120}{80}) \times (27.0 + 273.15)] K = 450.22 K = {177.07}^{\circ} C \approx 177^{\circ} C .$

Question 14

The ideal gas law, which relates the P, V and temperature T of an ideal gas in a closed box, given as PV = nRT, can be manipulated to construct an effective thermometer. A "constant volume gas thermometer” uses a sample of gas (commonly, nitrogen or helium) in a closed chamber, thus fixing the volume V. The temperature is then measured by reading off the corresponding pressure P. At the freezing point of water, it is seen that a gas thermometer records a pressure of 0.9 x 10⁵ Pa; it also records a pressure of 1.2 x 10⁵ Pa at the boiling point of water. What pressure will you find in the gas chamber at a room temperature of 23⁰C?

0.81 x 10⁵ Pa

1.34 x 10⁵ Pa

0.64 x 10⁵ Pa

0.97 x 10⁵ Pa.

SOLUTION

Solution : D

For an ideal gas at constant V,

P ∝ T

(for some proportionality constant 'C').

Let (T_F, P_F) and (T_B,P_B) be the (T,P) values for the freezing and the boiling points of water respectively. We can write -

(1)

Subtracting,

(2)

From relation (1) and (2), we can write -

(3)

Comparing (2) and (3),

Question 15

The resistance of pure platinum increases linearly with temperature (over a small range). We can use this property to measure temperature of different bodies, relative to a scale.

If we scale the thermometer such that a resistance of 80 Ω denotes $0^{o} C$ and 90 Ω denotes $100^{o} C$ , what temperature will a resistance of 86 Ω denote?

40⁰C

50⁰C

60⁰C

80⁰C

SOLUTION

Solution : C

A "linear” relation between two quantities, say, Q₁ and Q₂, means the following
$Q_{1} = m Q_{2} + c$
where m and c are constants. This follows from the fact that for a straight line on a plane, y and x are related as,
$y = m x + c$

$m$ being the slope and $c$ the y-intercept respectively.

In our case, this should be true between the temperature and the resistance at that temperature,
$R_{θ} = α θ + β$
for some constants α & β.
Let's find out what these constants are from the information given. We can write
$80 Ω = α \times 0^{o} C + β$
$\Rightarrow β = 80 Ω$
$90 Ω = α \times 100^{o} C + β$
$\Rightarrow α = \frac{90 - 80}{100} = 0.1 Ω /^{o} C$

Using these values, the temperature( $θ$ ) can be written as
$86 = 0.1 \times θ + 80$
$\Rightarrow θ = 60^{o} C$

Question 16

A glass of water at room temperature and a hot mug of coffee are placed on two tables inside a closed room. As the coffee cools down, what would happen to the temperature of the water?

Stays absolutely unaffected, since it is on another table.

The coffee and water cools down with time.

The water gets slightly warmer after taking heat from the surrounding air.

The mug of coffee takes out the heat from the coffee but does not transfer it to the surrounding.

SOLUTION

Solution : C

Let us assume the room is perfectly closed and no heat is being gained from or lost to the environment outside.

Here we have three bodies thermally interacting inside the room - the glass of water (initially at room temperature of 23 degree Celsius), the air in the room, and the hot mug of coffee.
As the coffee cools down, it slightly increases the hotness of the air in the room. Thus, the room temperature itself increases after some time. This means the glass of water is now slightly cooler than the air around it.

Due to this temperature difference, the water will get slightly warmer until it has the same temperature as the air through which exchange of heat occur.

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Question 16

SOLUTION

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