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Physics NEET Questions — Free MCQ Practice

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An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?

A.

$\frac{1}{2} m v^{3}$

B.

$m v^{3}$

C.

$\frac{1}{2} m v^{2}$

D.

$\frac{1}{2} m^{2} v^{2}$

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Explanation

Let m is the mass per unit length then rate of mass per sec= $\frac{m x}{t} = m v$

Rate of KE= $\frac{1}{2} (m v) v^{2} = \frac{1}{2} m v^{3}$

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Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g / cm³. If the mass of the other is 48 g, its density in g / cm³ is

A.

$\frac{4}{3}$

B.

$\frac{3}{2}$

C.

3

D.

5

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Explanation

(c) Apparent weight = $V (ρ - σ) g = \frac{m}{ρ} (ρ - σ) g$
where m = mass of the body,
$ρ$ = density of the body
$σ$ = density of water
If two bodies are in equilibrium then their apparent weight must be equal.

$∴ \frac{m_{1}}{ρ_{1}} (ρ_{1} - σ) = \frac{m_{2}}{ρ_{2}} (ρ_{2} - σ) \Rightarrow \frac{36}{9} (9 - 1) = \frac{48}{ρ_{2}} (ρ_{2} - 1)$

By solving we get $ρ_{2} = 3 .$

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An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm³ of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm³)

A.

350 cm³

B.

300 cm³

C.

250 cm³

D.

22 cm³

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The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. Ratio of density of mercury to that of air is $10^{4}$ . The height of the hill is

A.

250 m

B.

2.5 km

C.

1.25 km

D.

750 m

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Explanation

(b) Difference of pressure between sea level and the top of hill
$∆ P = (h_{1} - h_{2}) \times ρ_{Hg} \times g = (75 - 50) \times 10^{- 2} \times ρ_{Hg} \times g$ …(i)
and pressure difference due to h meter of air
$∆ P = h \times ρ_{air} \times g$ …(ii)
By equating (i) and (ii) we get

$h \times ρ_{air} \times g = (75 - 50) \times 10^{- 2} \times ρ_{Hg} \times g$

$∴ h = 25 \times 10^{- 2} (\frac{ρ_{Hg}}{ρ_{air}}) = 25 \times 10^{- 2} \times 10^{4} = 2500 m$

Height of the hill = 2.5 km.

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Equal masses of water and a liquid of relative density 2 are mixed together, then the mixture has a density of

A.

2/3

B.

4/3

C.

3/2

D.

3

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Explanation

(b) If two liquid of equal masses and different densities are mixed together then density of mixture

$ρ = \frac{2 ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}} = \frac{2 \times 1 \times 2}{1 + 2} = \frac{4}{3}$

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A body of density $d_{1}$ is counterpoised by Mg of weights of density $d_{2}$ in air of density d. Then the true mass of the body is

A.

M

B.

$M (1 - \frac{d}{d_{2}})$

C.

$M (1 - \frac{d}{d_{1}})$

D.

$\frac{M (1 - d / d_{2})}{(1 - d / d_{1})}$

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Explanation

(d) Let $M_{0}$ =mass of the body in the vacuum.
Apparent weight of the body in air = Apparent weight of standard weights in air
$\Rightarrow$ Actual weight – upthrust due to displaced air = Actual weight – upthrust due to displaced air

$\Rightarrow M_{0} g - (V_{0}) dg = Mg - (V) dg \Rightarrow M_{0} g - (\frac{M_{0}}{d_{1}}) dg = Mg - (\frac{M}{d_{2}}) dg \Rightarrow M_{0} = \frac{M [1 - \frac{d}{d^{2}}]}{[1 - \frac{d}{d_{1}}]}$

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The value of g at a place decreases by 2%. The barometric height of mercury

A.

Increases by 2%

B.

Decreases by 2%

C.

Remains unchanged

D.

Sometimes increases and sometimes decreases

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Explanation

(a) $h = \frac{P}{ρg} ∴ h \propto \frac{1}{g} (P and ρ are constant)$

If value of g decreased by 2% then h will increase by 2%.

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A barometer kept in a stationary elevator reads 76 cm. If the elevator starts accelerating up, the reading will be

A.

Zero

B.

Equal to 76 cm

C.

More than 76 cm

D.

Less than 76 cm

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Explanation

(d) $h = \frac{P}{ρg} ∴ h \propto \frac{1}{g}$ . If lift moves upward with some acceleration then effective g increases. So the value of h decreases i.e. reading will be less than 76 cm.

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A barometer tube reads 76 cm of mercury. If the tube is gradually inclined at an angle of 60^o with vertical, keeping the open end immersed in the mercury reservoir, the length of the mercury column will be

A.

152 cm

B.

76 cm

C.

38 cm

D.

$38 \sqrt{3}$ cm

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A triangular lamina of area A and height h is immersed in a liquid of density $ρ$ in a vertical plane with its base on the surface of the liquid. The thrust on the lamina is

A.

$\frac{1}{2} Aρgh$

B.

$\frac{1}{3} Aρgh$

C.

$\frac{1}{6} Aρgh$

D.

$\frac{2}{3} Aρgh$

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Explanation

(b) Thrust on lamina = pressure at centroid × Area

$= \frac{hρg}{3} \times A = \frac{1}{3} Aρgh$

Physics MCQs for NEET — Practice Questions with Answers

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