163170
How much work will be done in increasing the diameter of a soap bubble from \(2 \mathrm{~cm}\) to \(5 \mathrm{~cm}\). Surface tension of soap solution is \(3.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) : [RBQ]
1 \(4.86 \times 10^{-4} \mathrm{~J}\)
2 \(9.99 \times 10^{-4} \mathrm{~J}\)
3 \(3.96 \times 10^{-4} \mathrm{~J}\)
4 \(7.94 \times 10^{-4} \mathrm{~J}\)
Explanation:
Soap bubble has two surface, Hence , \(\mathrm{W}=\mathrm{T} \Delta \mathrm{A}\) Here, \(\Delta A=2\left\) \(=1.32 \times 10^{-2} \mathrm{~m}^2\) \(\therefore W =\left(3.0 \times 10^{-2}\right)\left(1.32 \times 10^{-2}\right) \mathrm{J}\) \(=3.96 \times 10^{-4} \mathrm{~J}\)
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 \(\mathrm{P}_1=\frac{2 \mathrm{~T}}{\mathrm{R}} ; \quad \mathrm{P}_2=\frac{2 \mathrm{~T}}{\mathrm{R}^{\prime}}\) i.e., \(P_1=\frac{2 T}{R}=4 \times \frac{2 T}{R^{\prime}}\) or \(\frac{2 T}{R}=\frac{2 T}{\left(R^{\prime} / 4\right)}\) \(\therefore \mathrm{R}^{\prime}=4 \mathrm{R}\) or \(\mathrm{R}=\mathrm{R}^{\prime} / 4\) \(\mathrm{m}_1=\frac{4}{3} \pi \mathrm{R}^3 \rho\) \(m_2=\frac{4}{3} \pi(4 R)^3 \rho=64 m_1\) \(\therefore \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{1}{64}\) or \(1: 64\)
NCERT-XI-I-196
6 RBTS PAPER
163172
Water rises in a capillary tube to a height of \(2.0 \mathrm{~cm}\). In another capillary tube whose radius is one third of it, how much the water will rise : [RBQ]
1 \(6.0 \mathrm{~cm}\)
2 \(10 \mathrm{~cm}\)
3 \(15 \mathrm{~cm}\)
4 \(09 \mathrm{~cm}\)
Explanation:
\( h=\frac{2 T \cos \theta}{r \rho g}\) \(\therefore \quad h r=\frac{2 T \cos \theta}{\rho g}=\text { constant }\) \(\therefore \quad h_1 r_1=h_2 r_2\) substituting the values \( h_2=(2.0)(3)\) \(=6.0 \mathrm{~cm}\)
NCERT-XI-I-197
6 RBTS PAPER
163173
A wire of area of cross-section \(10^{-6} \mathrm{~m}^2\) is increased in length by \(0.1 \%\). The tension produced is \(1000 \mathrm{~N}\). The Young's modulus of wire is [RBQ]
NEET Test Series from KOTA - 10 Papers In MS WORD
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6 RBTS PAPER
163170
How much work will be done in increasing the diameter of a soap bubble from \(2 \mathrm{~cm}\) to \(5 \mathrm{~cm}\). Surface tension of soap solution is \(3.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) : [RBQ]
1 \(4.86 \times 10^{-4} \mathrm{~J}\)
2 \(9.99 \times 10^{-4} \mathrm{~J}\)
3 \(3.96 \times 10^{-4} \mathrm{~J}\)
4 \(7.94 \times 10^{-4} \mathrm{~J}\)
Explanation:
Soap bubble has two surface, Hence , \(\mathrm{W}=\mathrm{T} \Delta \mathrm{A}\) Here, \(\Delta A=2\left\) \(=1.32 \times 10^{-2} \mathrm{~m}^2\) \(\therefore W =\left(3.0 \times 10^{-2}\right)\left(1.32 \times 10^{-2}\right) \mathrm{J}\) \(=3.96 \times 10^{-4} \mathrm{~J}\)
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 \(\mathrm{P}_1=\frac{2 \mathrm{~T}}{\mathrm{R}} ; \quad \mathrm{P}_2=\frac{2 \mathrm{~T}}{\mathrm{R}^{\prime}}\) i.e., \(P_1=\frac{2 T}{R}=4 \times \frac{2 T}{R^{\prime}}\) or \(\frac{2 T}{R}=\frac{2 T}{\left(R^{\prime} / 4\right)}\) \(\therefore \mathrm{R}^{\prime}=4 \mathrm{R}\) or \(\mathrm{R}=\mathrm{R}^{\prime} / 4\) \(\mathrm{m}_1=\frac{4}{3} \pi \mathrm{R}^3 \rho\) \(m_2=\frac{4}{3} \pi(4 R)^3 \rho=64 m_1\) \(\therefore \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{1}{64}\) or \(1: 64\)
NCERT-XI-I-196
6 RBTS PAPER
163172
Water rises in a capillary tube to a height of \(2.0 \mathrm{~cm}\). In another capillary tube whose radius is one third of it, how much the water will rise : [RBQ]
1 \(6.0 \mathrm{~cm}\)
2 \(10 \mathrm{~cm}\)
3 \(15 \mathrm{~cm}\)
4 \(09 \mathrm{~cm}\)
Explanation:
\( h=\frac{2 T \cos \theta}{r \rho g}\) \(\therefore \quad h r=\frac{2 T \cos \theta}{\rho g}=\text { constant }\) \(\therefore \quad h_1 r_1=h_2 r_2\) substituting the values \( h_2=(2.0)(3)\) \(=6.0 \mathrm{~cm}\)
NCERT-XI-I-197
6 RBTS PAPER
163173
A wire of area of cross-section \(10^{-6} \mathrm{~m}^2\) is increased in length by \(0.1 \%\). The tension produced is \(1000 \mathrm{~N}\). The Young's modulus of wire is [RBQ]
163170
How much work will be done in increasing the diameter of a soap bubble from \(2 \mathrm{~cm}\) to \(5 \mathrm{~cm}\). Surface tension of soap solution is \(3.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) : [RBQ]
1 \(4.86 \times 10^{-4} \mathrm{~J}\)
2 \(9.99 \times 10^{-4} \mathrm{~J}\)
3 \(3.96 \times 10^{-4} \mathrm{~J}\)
4 \(7.94 \times 10^{-4} \mathrm{~J}\)
Explanation:
Soap bubble has two surface, Hence , \(\mathrm{W}=\mathrm{T} \Delta \mathrm{A}\) Here, \(\Delta A=2\left\) \(=1.32 \times 10^{-2} \mathrm{~m}^2\) \(\therefore W =\left(3.0 \times 10^{-2}\right)\left(1.32 \times 10^{-2}\right) \mathrm{J}\) \(=3.96 \times 10^{-4} \mathrm{~J}\)
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 \(\mathrm{P}_1=\frac{2 \mathrm{~T}}{\mathrm{R}} ; \quad \mathrm{P}_2=\frac{2 \mathrm{~T}}{\mathrm{R}^{\prime}}\) i.e., \(P_1=\frac{2 T}{R}=4 \times \frac{2 T}{R^{\prime}}\) or \(\frac{2 T}{R}=\frac{2 T}{\left(R^{\prime} / 4\right)}\) \(\therefore \mathrm{R}^{\prime}=4 \mathrm{R}\) or \(\mathrm{R}=\mathrm{R}^{\prime} / 4\) \(\mathrm{m}_1=\frac{4}{3} \pi \mathrm{R}^3 \rho\) \(m_2=\frac{4}{3} \pi(4 R)^3 \rho=64 m_1\) \(\therefore \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{1}{64}\) or \(1: 64\)
NCERT-XI-I-196
6 RBTS PAPER
163172
Water rises in a capillary tube to a height of \(2.0 \mathrm{~cm}\). In another capillary tube whose radius is one third of it, how much the water will rise : [RBQ]
1 \(6.0 \mathrm{~cm}\)
2 \(10 \mathrm{~cm}\)
3 \(15 \mathrm{~cm}\)
4 \(09 \mathrm{~cm}\)
Explanation:
\( h=\frac{2 T \cos \theta}{r \rho g}\) \(\therefore \quad h r=\frac{2 T \cos \theta}{\rho g}=\text { constant }\) \(\therefore \quad h_1 r_1=h_2 r_2\) substituting the values \( h_2=(2.0)(3)\) \(=6.0 \mathrm{~cm}\)
NCERT-XI-I-197
6 RBTS PAPER
163173
A wire of area of cross-section \(10^{-6} \mathrm{~m}^2\) is increased in length by \(0.1 \%\). The tension produced is \(1000 \mathrm{~N}\). The Young's modulus of wire is [RBQ]
163170
How much work will be done in increasing the diameter of a soap bubble from \(2 \mathrm{~cm}\) to \(5 \mathrm{~cm}\). Surface tension of soap solution is \(3.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) : [RBQ]
1 \(4.86 \times 10^{-4} \mathrm{~J}\)
2 \(9.99 \times 10^{-4} \mathrm{~J}\)
3 \(3.96 \times 10^{-4} \mathrm{~J}\)
4 \(7.94 \times 10^{-4} \mathrm{~J}\)
Explanation:
Soap bubble has two surface, Hence , \(\mathrm{W}=\mathrm{T} \Delta \mathrm{A}\) Here, \(\Delta A=2\left\) \(=1.32 \times 10^{-2} \mathrm{~m}^2\) \(\therefore W =\left(3.0 \times 10^{-2}\right)\left(1.32 \times 10^{-2}\right) \mathrm{J}\) \(=3.96 \times 10^{-4} \mathrm{~J}\)
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 \(\mathrm{P}_1=\frac{2 \mathrm{~T}}{\mathrm{R}} ; \quad \mathrm{P}_2=\frac{2 \mathrm{~T}}{\mathrm{R}^{\prime}}\) i.e., \(P_1=\frac{2 T}{R}=4 \times \frac{2 T}{R^{\prime}}\) or \(\frac{2 T}{R}=\frac{2 T}{\left(R^{\prime} / 4\right)}\) \(\therefore \mathrm{R}^{\prime}=4 \mathrm{R}\) or \(\mathrm{R}=\mathrm{R}^{\prime} / 4\) \(\mathrm{m}_1=\frac{4}{3} \pi \mathrm{R}^3 \rho\) \(m_2=\frac{4}{3} \pi(4 R)^3 \rho=64 m_1\) \(\therefore \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{1}{64}\) or \(1: 64\)
NCERT-XI-I-196
6 RBTS PAPER
163172
Water rises in a capillary tube to a height of \(2.0 \mathrm{~cm}\). In another capillary tube whose radius is one third of it, how much the water will rise : [RBQ]
1 \(6.0 \mathrm{~cm}\)
2 \(10 \mathrm{~cm}\)
3 \(15 \mathrm{~cm}\)
4 \(09 \mathrm{~cm}\)
Explanation:
\( h=\frac{2 T \cos \theta}{r \rho g}\) \(\therefore \quad h r=\frac{2 T \cos \theta}{\rho g}=\text { constant }\) \(\therefore \quad h_1 r_1=h_2 r_2\) substituting the values \( h_2=(2.0)(3)\) \(=6.0 \mathrm{~cm}\)
NCERT-XI-I-197
6 RBTS PAPER
163173
A wire of area of cross-section \(10^{-6} \mathrm{~m}^2\) is increased in length by \(0.1 \%\). The tension produced is \(1000 \mathrm{~N}\). The Young's modulus of wire is [RBQ]