QBManager

6 RBTS PAPER

163170 How much work will be done in increasing the diameter of a soap bubble from $2 cm$ to $5 cm$ .
Surface tension of soap solution is $3.0 \times 10^{- 2} N / m$ : [RBQ]

1

4.86 \times 10^{- 4} J

2

9.99 \times 10^{- 4} J

3

3.96 \times 10^{- 4} J

4

7.94 \times 10^{- 4} J

Explanation:

Soap bubble has two surface, Hence ,
$W = T Δ A$
Here, $Missing or unrecognized delimiter for \left$
$= 1.32 \times 10^{- 2} m^{2}$
$∴ W = (3.0 \times 10^{- 2}) (1.32 \times 10^{- 2}) J$
$= 3.96 \times 10^{- 4} J$

NCERT-XI-I-196

6 RBTS PAPER

163171 The excess pressure inside a spherical drop of water is four times that of another drop. Then their respective mass ratio is : [RBQ]

1

1 : 16

2

8 : 1

3

1 : 4

4

1 : 64

Explanation:

Excess pressure inside the liquid drop $= \frac{2 T}{R}$ , where $T$ is surface tension and $R$ is its radius.
If $P_{1} = \frac{2 T}{R}; P_{2} = \frac{2 T}{R^{'}}$
i.e., $P_{1} = \frac{2 T}{R} = 4 \times \frac{2 T}{R^{'}}$
or $\frac{2 T}{R} = \frac{2 T}{(R^{'} / 4)}$
$∴ R^{'} = 4 R$ or $R = R^{'} / 4$
$m_{1} = \frac{4}{3} π R^{3} ρ$
$m_{2} = \frac{4}{3} π (4 R)^{3} ρ = 64 m_{1}$
$∴ \frac{m_{1}}{m_{2}} = \frac{1}{64}$ or $1 : 64$

NCERT-XI-I-196

6 RBTS PAPER

163172 Water rises in a capillary tube to a height of $2.0 cm$ . In another capillary tube whose radius is one third of it, how much the water will rise : [RBQ]

1

6.0 cm

2

10 cm

3

15 cm

4

09 cm

Explanation:

$h = \frac{2 T \cos θ}{r ρ g}$
$∴ h r = \frac{2 T \cos θ}{ρ g} = constant$
$∴ h_{1} r_{1} = h_{2} r_{2}$
substituting the values
$h_{2} = (2.0) (3)$
$= 6.0 cm$

NCERT-XI-I-197

6 RBTS PAPER

163173 A wire of area of cross-section $10^{- 6} m^{2}$ is increased in length by $0.1 %$ . The tension produced is $1000 N$ . The Young's modulus of wire is [RBQ]

1

10^{12} N / m^{2}

2

10^{11} N / m^{2}

3

10^{10} N / m^{2}

4

10^{9} N / m^{2}

Explanation:

$(\frac{r_{2}}{r_{1}} = \frac{1}{3})$
$Y = \frac{F L}{A Δ L} = \frac{1000 \times 100}{10^{- 6} \times 0.1} = 10^{12} N / m^{2}$

NCERT-XI-I-171

6 RBTS PAPER

163170 How much work will be done in increasing the diameter of a soap bubble from $2 cm$ to $5 cm$ .
Surface tension of soap solution is $3.0 \times 10^{- 2} N / m$ : [RBQ]

1

4.86 \times 10^{- 4} J

2

9.99 \times 10^{- 4} J

3

3.96 \times 10^{- 4} J

4

7.94 \times 10^{- 4} J

Explanation:

Soap bubble has two surface, Hence ,
$W = T Δ A$
Here, $Missing or unrecognized delimiter for \left$
$= 1.32 \times 10^{- 2} m^{2}$
$∴ W = (3.0 \times 10^{- 2}) (1.32 \times 10^{- 2}) J$
$= 3.96 \times 10^{- 4} J$

NCERT-XI-I-196

6 RBTS PAPER

163171 The excess pressure inside a spherical drop of water is four times that of another drop. Then their respective mass ratio is : [RBQ]

1

1 : 16

2

8 : 1

3

1 : 4

4

1 : 64

Explanation:

Excess pressure inside the liquid drop $= \frac{2 T}{R}$ , where $T$ is surface tension and $R$ is its radius.
If $P_{1} = \frac{2 T}{R}; P_{2} = \frac{2 T}{R^{'}}$
i.e., $P_{1} = \frac{2 T}{R} = 4 \times \frac{2 T}{R^{'}}$
or $\frac{2 T}{R} = \frac{2 T}{(R^{'} / 4)}$
$∴ R^{'} = 4 R$ or $R = R^{'} / 4$
$m_{1} = \frac{4}{3} π R^{3} ρ$
$m_{2} = \frac{4}{3} π (4 R)^{3} ρ = 64 m_{1}$
$∴ \frac{m_{1}}{m_{2}} = \frac{1}{64}$ or $1 : 64$

NCERT-XI-I-196

6 RBTS PAPER

163172 Water rises in a capillary tube to a height of $2.0 cm$ . In another capillary tube whose radius is one third of it, how much the water will rise : [RBQ]

1

6.0 cm

2

10 cm

3

15 cm

4

09 cm

Explanation:

$h = \frac{2 T \cos θ}{r ρ g}$
$∴ h r = \frac{2 T \cos θ}{ρ g} = constant$
$∴ h_{1} r_{1} = h_{2} r_{2}$
substituting the values
$h_{2} = (2.0) (3)$
$= 6.0 cm$

NCERT-XI-I-197

6 RBTS PAPER

163173 A wire of area of cross-section $10^{- 6} m^{2}$ is increased in length by $0.1 %$ . The tension produced is $1000 N$ . The Young's modulus of wire is [RBQ]

1

10^{12} N / m^{2}

2

10^{11} N / m^{2}

3

10^{10} N / m^{2}

4

10^{9} N / m^{2}

Explanation:

$(\frac{r_{2}}{r_{1}} = \frac{1}{3})$
$Y = \frac{F L}{A Δ L} = \frac{1000 \times 100}{10^{- 6} \times 0.1} = 10^{12} N / m^{2}$

NCERT-XI-I-171

6 RBTS PAPER

163170 How much work will be done in increasing the diameter of a soap bubble from $2 cm$ to $5 cm$ .
Surface tension of soap solution is $3.0 \times 10^{- 2} N / m$ : [RBQ]

1

4.86 \times 10^{- 4} J

2

9.99 \times 10^{- 4} J

3

3.96 \times 10^{- 4} J

4

7.94 \times 10^{- 4} J

Explanation:

Soap bubble has two surface, Hence ,
$W = T Δ A$
Here, $Missing or unrecognized delimiter for \left$
$= 1.32 \times 10^{- 2} m^{2}$
$∴ W = (3.0 \times 10^{- 2}) (1.32 \times 10^{- 2}) J$
$= 3.96 \times 10^{- 4} J$

NCERT-XI-I-196

6 RBTS PAPER

163171 The excess pressure inside a spherical drop of water is four times that of another drop. Then their respective mass ratio is : [RBQ]

1

1 : 16

2

8 : 1

3

1 : 4

4

1 : 64

Explanation:

Excess pressure inside the liquid drop $= \frac{2 T}{R}$ , where $T$ is surface tension and $R$ is its radius.
If $P_{1} = \frac{2 T}{R}; P_{2} = \frac{2 T}{R^{'}}$
i.e., $P_{1} = \frac{2 T}{R} = 4 \times \frac{2 T}{R^{'}}$
or $\frac{2 T}{R} = \frac{2 T}{(R^{'} / 4)}$
$∴ R^{'} = 4 R$ or $R = R^{'} / 4$
$m_{1} = \frac{4}{3} π R^{3} ρ$
$m_{2} = \frac{4}{3} π (4 R)^{3} ρ = 64 m_{1}$
$∴ \frac{m_{1}}{m_{2}} = \frac{1}{64}$ or $1 : 64$

NCERT-XI-I-196

6 RBTS PAPER

163172 Water rises in a capillary tube to a height of $2.0 cm$ . In another capillary tube whose radius is one third of it, how much the water will rise : [RBQ]

1

6.0 cm

2

10 cm

3

15 cm

4

09 cm

Explanation:

$h = \frac{2 T \cos θ}{r ρ g}$
$∴ h r = \frac{2 T \cos θ}{ρ g} = constant$
$∴ h_{1} r_{1} = h_{2} r_{2}$
substituting the values
$h_{2} = (2.0) (3)$
$= 6.0 cm$

NCERT-XI-I-197

6 RBTS PAPER

163173 A wire of area of cross-section $10^{- 6} m^{2}$ is increased in length by $0.1 %$ . The tension produced is $1000 N$ . The Young's modulus of wire is [RBQ]

1

10^{12} N / m^{2}

2

10^{11} N / m^{2}

3

10^{10} N / m^{2}

4

10^{9} N / m^{2}

Explanation:

$(\frac{r_{2}}{r_{1}} = \frac{1}{3})$
$Y = \frac{F L}{A Δ L} = \frac{1000 \times 100}{10^{- 6} \times 0.1} = 10^{12} N / m^{2}$

NCERT-XI-I-171

6 RBTS PAPER

163170 How much work will be done in increasing the diameter of a soap bubble from $2 cm$ to $5 cm$ .
Surface tension of soap solution is $3.0 \times 10^{- 2} N / m$ : [RBQ]

1

4.86 \times 10^{- 4} J

2

9.99 \times 10^{- 4} J

3

3.96 \times 10^{- 4} J

4

7.94 \times 10^{- 4} J

Explanation:

Soap bubble has two surface, Hence ,
$W = T Δ A$
Here, $Missing or unrecognized delimiter for \left$
$= 1.32 \times 10^{- 2} m^{2}$
$∴ W = (3.0 \times 10^{- 2}) (1.32 \times 10^{- 2}) J$
$= 3.96 \times 10^{- 4} J$

NCERT-XI-I-196

6 RBTS PAPER

163171 The excess pressure inside a spherical drop of water is four times that of another drop. Then their respective mass ratio is : [RBQ]

1

1 : 16

2

8 : 1

3

1 : 4

4

1 : 64

Explanation:

Excess pressure inside the liquid drop $= \frac{2 T}{R}$ , where $T$ is surface tension and $R$ is its radius.
If $P_{1} = \frac{2 T}{R}; P_{2} = \frac{2 T}{R^{'}}$
i.e., $P_{1} = \frac{2 T}{R} = 4 \times \frac{2 T}{R^{'}}$
or $\frac{2 T}{R} = \frac{2 T}{(R^{'} / 4)}$
$∴ R^{'} = 4 R$ or $R = R^{'} / 4$
$m_{1} = \frac{4}{3} π R^{3} ρ$
$m_{2} = \frac{4}{3} π (4 R)^{3} ρ = 64 m_{1}$
$∴ \frac{m_{1}}{m_{2}} = \frac{1}{64}$ or $1 : 64$

NCERT-XI-I-196

6 RBTS PAPER

163172 Water rises in a capillary tube to a height of $2.0 cm$ . In another capillary tube whose radius is one third of it, how much the water will rise : [RBQ]

1

6.0 cm

2

10 cm

3

15 cm

4

09 cm

Explanation:

$h = \frac{2 T \cos θ}{r ρ g}$
$∴ h r = \frac{2 T \cos θ}{ρ g} = constant$
$∴ h_{1} r_{1} = h_{2} r_{2}$
substituting the values
$h_{2} = (2.0) (3)$
$= 6.0 cm$

NCERT-XI-I-197

6 RBTS PAPER

163173 A wire of area of cross-section $10^{- 6} m^{2}$ is increased in length by $0.1 %$ . The tension produced is $1000 N$ . The Young's modulus of wire is [RBQ]

1

10^{12} N / m^{2}

2

10^{11} N / m^{2}

3

10^{10} N / m^{2}

4

10^{9} N / m^{2}

Explanation:

$(\frac{r_{2}}{r_{1}} = \frac{1}{3})$
$Y = \frac{F L}{A Δ L} = \frac{1000 \times 100}{10^{- 6} \times 0.1} = 10^{12} N / m^{2}$

NCERT-XI-I-171

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