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Mechanical Properties of Fluids

142807 A ring of radius $R$ is kept on water surface.
Surface tension of water is $T$ and mass is $m$ . What force required to lift the ring from water surface?

1

(mg + 2 T π R)

2

(2 mg + π R)

3

(3 mg + 2 π TR)

4

(mg + 4 T π R)

Explanation:

D
Force required to lift the ring,
External force $= mg + 2 \times T (2 π R)$
$= mg + 4 T π R$

AIIMS-27.05.2018(E)

Mechanical Properties of Fluids

142808 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

1

0.94 m

2

0.97 m

3

0.98 m

4

0.99 m

Explanation:

C Weight of the body $=$ weight of water displaced
$W_{b} = W_{w}$
$V_{b} \times ρ_{b} \times g = V_{w} \times ρ_{w} \times g$
$\frac{2}{3} π (R^{3} - r^{3}) \times 2 \times 10^{4} = \frac{2}{3} π R^{3} \times 1.2 \times 10^{3}$
$20 (R^{3} - r^{3}) = 1.2 R^{3}$
$r^{3} = \frac{18.8}{20} R^{3}$
$r = 0.98 R$
$2 r = 0.98 (2 R) = 0.98 m$
$(∵ 2 R = 1 m)$
Hence, inner diameter is $0.98 m$ .

AIIMS-27.05.2018(E)

Mechanical Properties of Fluids

142812 Work of 3.0 $\times 10^{- 4}$ joule is required to be done in increasing the size of a soap film from $10 cm \times$ $6 cm$ to $10 cm \times 11 cm$ . The surface tension of the film is

1

5 \times 10^{- 2} N / m

2

3 \times 10^{- 2} N / m

3

1.5 \times 10^{- 2} N / m

4

1.2 \times 10^{- 2} N / m

Explanation:

B Work done $= 3.0 \times 10^{- 4} J$
Increased Area $= (10 \times 11 - 10 \times 6) = 50 {cm}^{2}$
Film has two sides,
$∴$ Total increased area $= 50 \times 2 = 100 {cm}^{2}$
$Surface tension = \frac{Work done}{increase surface area}$
$= \frac{3 \times 10^{- 4}}{100 \times 10^{- 4}} = 0.03 N / m$
$= 3 \times 10^{- 2} N / m$

AIIMS-2007

Mechanical Properties of Fluids

142816 A mercury drop of radius $1 cm$ is sprayed into $10^{6}$ drops of equal size. The energy expressed in joule is (surface tension of Mercury is $460 \times 10^{- 3}$ $N / m)$

1 0.057

2 5.7

3

5.7 \times 10^{- 4}

4

5.7 \times 10^{- 6}

Explanation:

A Given,
$R = 1 cm = 10^{- 2} m$
$n = 10^{6}$
$T = 460 \times 10^{- 3} N / m$
We know that,
Work done in breaking mercury drop in small droplets-
$W = T \times 4 π R^{2} (n^{1 / 3} - 1)$
$W = 460 \times 10^{- 3} \times 4 \times 3.14 \times {(10^{- 2})}^{2} [{(10^{6})}^{1 / 3} - 1]$
$W = 0.057$ joule

AIIMS-26.05.2019(M) Shift-1

Mechanical Properties of Fluids

142807 A ring of radius $R$ is kept on water surface.
Surface tension of water is $T$ and mass is $m$ . What force required to lift the ring from water surface?

1

(mg + 2 T π R)

2

(2 mg + π R)

3

(3 mg + 2 π TR)

4

(mg + 4 T π R)

Explanation:

D
Force required to lift the ring,
External force $= mg + 2 \times T (2 π R)$
$= mg + 4 T π R$

AIIMS-27.05.2018(E)

Mechanical Properties of Fluids

142808 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

1

0.94 m

2

0.97 m

3

0.98 m

4

0.99 m

Explanation:

C Weight of the body $=$ weight of water displaced
$W_{b} = W_{w}$
$V_{b} \times ρ_{b} \times g = V_{w} \times ρ_{w} \times g$
$\frac{2}{3} π (R^{3} - r^{3}) \times 2 \times 10^{4} = \frac{2}{3} π R^{3} \times 1.2 \times 10^{3}$
$20 (R^{3} - r^{3}) = 1.2 R^{3}$
$r^{3} = \frac{18.8}{20} R^{3}$
$r = 0.98 R$
$2 r = 0.98 (2 R) = 0.98 m$
$(∵ 2 R = 1 m)$
Hence, inner diameter is $0.98 m$ .

AIIMS-27.05.2018(E)

Mechanical Properties of Fluids

142812 Work of 3.0 $\times 10^{- 4}$ joule is required to be done in increasing the size of a soap film from $10 cm \times$ $6 cm$ to $10 cm \times 11 cm$ . The surface tension of the film is

1

5 \times 10^{- 2} N / m

2

3 \times 10^{- 2} N / m

3

1.5 \times 10^{- 2} N / m

4

1.2 \times 10^{- 2} N / m

Explanation:

B Work done $= 3.0 \times 10^{- 4} J$
Increased Area $= (10 \times 11 - 10 \times 6) = 50 {cm}^{2}$
Film has two sides,
$∴$ Total increased area $= 50 \times 2 = 100 {cm}^{2}$
$Surface tension = \frac{Work done}{increase surface area}$
$= \frac{3 \times 10^{- 4}}{100 \times 10^{- 4}} = 0.03 N / m$
$= 3 \times 10^{- 2} N / m$

AIIMS-2007

Mechanical Properties of Fluids

142816 A mercury drop of radius $1 cm$ is sprayed into $10^{6}$ drops of equal size. The energy expressed in joule is (surface tension of Mercury is $460 \times 10^{- 3}$ $N / m)$

1 0.057

2 5.7

3

5.7 \times 10^{- 4}

4

5.7 \times 10^{- 6}

Explanation:

A Given,
$R = 1 cm = 10^{- 2} m$
$n = 10^{6}$
$T = 460 \times 10^{- 3} N / m$
We know that,
Work done in breaking mercury drop in small droplets-
$W = T \times 4 π R^{2} (n^{1 / 3} - 1)$
$W = 460 \times 10^{- 3} \times 4 \times 3.14 \times {(10^{- 2})}^{2} [{(10^{6})}^{1 / 3} - 1]$
$W = 0.057$ joule

AIIMS-26.05.2019(M) Shift-1

Mechanical Properties of Fluids

142807 A ring of radius $R$ is kept on water surface.
Surface tension of water is $T$ and mass is $m$ . What force required to lift the ring from water surface?

1

(mg + 2 T π R)

2

(2 mg + π R)

3

(3 mg + 2 π TR)

4

(mg + 4 T π R)

Explanation:

D
Force required to lift the ring,
External force $= mg + 2 \times T (2 π R)$
$= mg + 4 T π R$

AIIMS-27.05.2018(E)

Mechanical Properties of Fluids

142808 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

1

0.94 m

2

0.97 m

3

0.98 m

4

0.99 m

Explanation:

C Weight of the body $=$ weight of water displaced
$W_{b} = W_{w}$
$V_{b} \times ρ_{b} \times g = V_{w} \times ρ_{w} \times g$
$\frac{2}{3} π (R^{3} - r^{3}) \times 2 \times 10^{4} = \frac{2}{3} π R^{3} \times 1.2 \times 10^{3}$
$20 (R^{3} - r^{3}) = 1.2 R^{3}$
$r^{3} = \frac{18.8}{20} R^{3}$
$r = 0.98 R$
$2 r = 0.98 (2 R) = 0.98 m$
$(∵ 2 R = 1 m)$
Hence, inner diameter is $0.98 m$ .

AIIMS-27.05.2018(E)

Mechanical Properties of Fluids

142812 Work of 3.0 $\times 10^{- 4}$ joule is required to be done in increasing the size of a soap film from $10 cm \times$ $6 cm$ to $10 cm \times 11 cm$ . The surface tension of the film is

1

5 \times 10^{- 2} N / m

2

3 \times 10^{- 2} N / m

3

1.5 \times 10^{- 2} N / m

4

1.2 \times 10^{- 2} N / m

Explanation:

B Work done $= 3.0 \times 10^{- 4} J$
Increased Area $= (10 \times 11 - 10 \times 6) = 50 {cm}^{2}$
Film has two sides,
$∴$ Total increased area $= 50 \times 2 = 100 {cm}^{2}$
$Surface tension = \frac{Work done}{increase surface area}$
$= \frac{3 \times 10^{- 4}}{100 \times 10^{- 4}} = 0.03 N / m$
$= 3 \times 10^{- 2} N / m$

AIIMS-2007

Mechanical Properties of Fluids

142816 A mercury drop of radius $1 cm$ is sprayed into $10^{6}$ drops of equal size. The energy expressed in joule is (surface tension of Mercury is $460 \times 10^{- 3}$ $N / m)$

1 0.057

2 5.7

3

5.7 \times 10^{- 4}

4

5.7 \times 10^{- 6}

Explanation:

A Given,
$R = 1 cm = 10^{- 2} m$
$n = 10^{6}$
$T = 460 \times 10^{- 3} N / m$
We know that,
Work done in breaking mercury drop in small droplets-
$W = T \times 4 π R^{2} (n^{1 / 3} - 1)$
$W = 460 \times 10^{- 3} \times 4 \times 3.14 \times {(10^{- 2})}^{2} [{(10^{6})}^{1 / 3} - 1]$
$W = 0.057$ joule

AIIMS-26.05.2019(M) Shift-1

Mechanical Properties of Fluids

142807 A ring of radius $R$ is kept on water surface.
Surface tension of water is $T$ and mass is $m$ . What force required to lift the ring from water surface?

1

(mg + 2 T π R)

2

(2 mg + π R)

3

(3 mg + 2 π TR)

4

(mg + 4 T π R)

Explanation:

D
Force required to lift the ring,
External force $= mg + 2 \times T (2 π R)$
$= mg + 4 T π R$

AIIMS-27.05.2018(E)

Mechanical Properties of Fluids

142808 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

1

0.94 m

2

0.97 m

3

0.98 m

4

0.99 m

Explanation:

C Weight of the body $=$ weight of water displaced
$W_{b} = W_{w}$
$V_{b} \times ρ_{b} \times g = V_{w} \times ρ_{w} \times g$
$\frac{2}{3} π (R^{3} - r^{3}) \times 2 \times 10^{4} = \frac{2}{3} π R^{3} \times 1.2 \times 10^{3}$
$20 (R^{3} - r^{3}) = 1.2 R^{3}$
$r^{3} = \frac{18.8}{20} R^{3}$
$r = 0.98 R$
$2 r = 0.98 (2 R) = 0.98 m$
$(∵ 2 R = 1 m)$
Hence, inner diameter is $0.98 m$ .

AIIMS-27.05.2018(E)

Mechanical Properties of Fluids

142812 Work of 3.0 $\times 10^{- 4}$ joule is required to be done in increasing the size of a soap film from $10 cm \times$ $6 cm$ to $10 cm \times 11 cm$ . The surface tension of the film is

1

5 \times 10^{- 2} N / m

2

3 \times 10^{- 2} N / m

3

1.5 \times 10^{- 2} N / m

4

1.2 \times 10^{- 2} N / m

Explanation:

B Work done $= 3.0 \times 10^{- 4} J$
Increased Area $= (10 \times 11 - 10 \times 6) = 50 {cm}^{2}$
Film has two sides,
$∴$ Total increased area $= 50 \times 2 = 100 {cm}^{2}$
$Surface tension = \frac{Work done}{increase surface area}$
$= \frac{3 \times 10^{- 4}}{100 \times 10^{- 4}} = 0.03 N / m$
$= 3 \times 10^{- 2} N / m$

AIIMS-2007

Mechanical Properties of Fluids

142816 A mercury drop of radius $1 cm$ is sprayed into $10^{6}$ drops of equal size. The energy expressed in joule is (surface tension of Mercury is $460 \times 10^{- 3}$ $N / m)$

1 0.057

2 5.7

3

5.7 \times 10^{- 4}

4

5.7 \times 10^{- 6}

Explanation:

A Given,
$R = 1 cm = 10^{- 2} m$
$n = 10^{6}$
$T = 460 \times 10^{- 3} N / m$
We know that,
Work done in breaking mercury drop in small droplets-
$W = T \times 4 π R^{2} (n^{1 / 3} - 1)$
$W = 460 \times 10^{- 3} \times 4 \times 3.14 \times {(10^{- 2})}^{2} [{(10^{6})}^{1 / 3} - 1]$
$W = 0.057$ joule

AIIMS-26.05.2019(M) Shift-1