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Physics NEET Questions — Free MCQ Practice

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If two liquids of same masses but densities $ρ_{1}$ and $ρ_{2}$ respectively are mixed, then density of mixture is given by

A.

$ρ = \frac{ρ_{1} + ρ_{2}}{2}$

B.

$ρ = \frac{ρ_{1} + ρ_{2}}{2 ρ_{1} ρ_{2}}$

C.

$ρ = \frac{2 ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}$

D.

$ρ = \frac{ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}$

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Explanation

(c) $ρ = \frac{Total mass}{Total volume} = \frac{2 m}{V_{1} + V_{2}} = \frac{2 m}{m (\frac{1}{ρ_{1}} + \frac{1}{ρ_{2}})}$

$∴ ρ = \frac{2 ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}$

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If two liquids of same volume but different densities $ρ_{1}$ and $ρ_{2}$ are mixed, then density of mixture is given by

A.

$ρ = \frac{ρ_{1} + ρ_{2}}{2}$

B.

$ρ = \frac{ρ_{1} + ρ_{2}}{2 ρ_{1} ρ_{2}}$

C.

$ρ = \frac{2 ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}$

D.

$ρ = \frac{ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}$

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Explanation

(a) $ρ = \frac{Total mass}{Total volume} = \frac{m_{1} + m_{2}}{2 V} = \frac{V (ρ_{1} + ρ_{2})}{2 V} = \frac{ρ_{1} + ρ_{2}}{2}$

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With rise in temperature, density of a given body changes according to one of the following relations

A.

$ρ = ρ_{0} [1 + γ dθ]$

B.

$ρ = ρ_{0} [1 - γ dθ]$

C.

$ρ = ρ_{0} γ dθ$

D.

$ρ = ρ_{0} / γ dθ$

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Explanation

(b) Since, with increase in temperature, volume of given body increases, while mass remains constant so that density will decrease.

i.e. $\frac{ρ}{ρ_{0}} = \frac{m / V}{m / V_{0}} = \frac{V_{0}}{V} = \frac{V}{V_{0} (1 + γ ∆ θ)} = (1 - γ ∆ θ) ∴ ρ = ρ_{0} (1 - γ ∆ θ)$

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An ice berg of density 900 Kg/ $m^{3}$ is floating in water of density 1000 Kg/ $m^{3}$ . The percentage of volume of ice-cube outside the water is

A.

20%

B.

35%

C.

10%

D.

25%

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Explanation

(c) Let the total volume of ice-berg is V and its density is $ρ$ . If this ice-berg floats in water with volume $V_{i n}$

inside it then $V_{in} σg = Vρg \Rightarrow V_{in} = (\frac{ρ}{σ}) V$

or $V_{out} = V - V_{in} = (\frac{σ - ρ}{σ}) V \Rightarrow \frac{V_{out}}{V} = (\frac{σ - ρ}{σ}) = \frac{1000 - 900}{1000} = \frac{1}{10} ∴ V_{out} = 10 % of V$

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A hemispherical bowl just floats without sinking in a liquid of density $1.2 \times 10^{3} kg / m^{3}$ . If outer diameter and the density of the bowl are 1 m and $2 \times 10^{4} kg / m^{3}$ respectively, then the inner diameter of the bowl will be

A.

0.94 m

B.

0.97 m

C.

0.98 m

D.

0.99 m

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Explanation

(c) Weight of the bowl = mg

= $Vρg = \frac{2}{3} π [{(\frac{D}{2})}^{3} - {(\frac{d}{2})}^{3}] ρg$

where D = Outer diameter ,

d = Inner diameter

$ρ$ = Density of bowl

Weight of the liquid displaced by the bowl

$= Vσg = \frac{2}{3} π {(\frac{D}{2})}^{3} σg$

where $σ$ is the density of the liquid

For the floatation $\frac{2}{3} π {(\frac{D}{2})}^{2} σg = \frac{2}{3} π [{(\frac{D}{2})}^{3} - {(\frac{d}{2})}^{3}] ρg \Rightarrow {(\frac{1}{2})}^{3} \times 1.2 \times 10^{3} = [{(\frac{1}{2})}^{3} - {(\frac{d}{2})}^{3}] 2 \times 10^{4}$

By solving we get d = 0.98 m.

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A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The specific gravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of sawdust will be

A.

8

B.

4

C.

3

D.

Zero

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Explanation

(b) Let specific gravities of concrete and saw dust are $ρ_{1}$ and $ρ_{2}$ respectively.
According to principle of floatation weight of whole sphere = upthrust on the sphere

$\frac{4}{3} π (R^{3} - r^{3}) ρ_{1} g + \frac{4}{3} {πr}^{3} ρ_{2} g = \frac{4}{3} {πR}^{3} \times 1 \times g \Rightarrow R^{3} ρ_{1} - r^{3} ρ_{1} + r^{3} ρ_{2} = R^{3} \Rightarrow R^{3} (ρ_{1} - 1) = r^{3} (ρ_{1} - ρ_{2}) \Rightarrow \frac{R^{3}}{r^{3}} = \frac{ρ_{1} - ρ_{2}}{ρ_{1} - 1} \Rightarrow \frac{R^{3} - r^{3}}{r^{3}} = \frac{ρ_{1} - ρ_{2} - ρ_{1} + 1}{ρ_{1} - 1} \Rightarrow \frac{(R^{3} - r^{3}) ρ_{1}}{r^{3} ρ_{2}} = (\frac{1 - ρ_{2}}{ρ_{1} - 1}) \frac{ρ_{1}}{ρ_{2}} \Rightarrow \frac{Mass of concrete}{Mass of saw dust} = (\frac{1 - 0.3}{2.4 - 1}) \times \frac{2.4}{0.3} = 4$

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A metallic block of density 5 gm ${cm}^{- 3}$ and having dimensions 5 cm × 5 cm × 5 cm is weighed in water. Its apparent weight will be

A.

5 × 5 × 5 × 5 gf

B.

4 × 4 × 4 × 4 gf

C.

5 × 4 × 4 × 4 gf

D.

4 × 5 × 5 × 5 gf

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Explanation

(d) Apparent weight

$= V (ρ - σ) g = l \times b \times h \times (5 - 1) \times g = 5 \times 5 \times 5 \times 4 \times g Dyne = 4 \times 5 \times 5 \times 5 gf$

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A silver ingot weighing 2.1 kg is held by a string so as to be completely immersed in a liquid of relative density 0.8. The relative density of silver is 10.5. The tension in the string in kg-wt is

A.

1.6

B.

1.94

C.

3.1

D.

5.25

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Explanation

(b) Apparent weight

$= V (ρ - σ) g = \frac{M}{ρ} (ρ - σ) g$

$= M (1 - \frac{σ}{ρ}) g = 2.1 (1 - \frac{0.8}{10.5}) g = 1.94 g N = 1.94 kg - wt$

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Two solids A and B float in water. It is observed that A floats with half its volume immersed and B floats with 2/3 of its volume immersed. Compare the densities of A and B

A.

4:3

B.

2:3

C.

3:4

D.

1:3

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Explanation

(c) If two different bodies A and B are floating in the same liquid then

$\frac{ρ_{A}}{ρ_{B}} = \frac{(f_{in})_{A}}{(f_{in})_{B}} = \frac{1 / 2}{2 / 3} = \frac{3}{4}$

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The fraction of a floating object of volume $V_{0}$ and density $d_{0}$ above the surface of a liquid of density d will be

A.

$\frac{d_{0}}{d}$

B.

$\frac{{dd}_{0}}{d + d_{0}}$

C.

$\frac{d - d_{0}}{d}$

D.

$\frac{{dd}_{0}}{d - d_{0}}$

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Explanation

(c) For the floatation $V_{0} d_{0} g = V_{in} dg$

$V_{in} = V_{0} \frac{d_{0}}{d}$

$∴ V_{out} = V_{0} - V_{in} = V_{0} - V_{0} \frac{d_{0}}{d} = V_{0} [\frac{d - d_{0}}{d}] \Rightarrow \frac{V_{out}}{V_{0}} = \frac{d - d_{0}}{d}$

Physics MCQs for NEET — Practice Questions with Answers

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