6 RBTS PAPER(PHYSICS)
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
6 RBTS PAPER

163182 The viscous force acting on a body falling under gravity in viscous fluid will be : [RBQ]

1 \(\frac{6 \pi V}{\eta r}\)
2 \(\frac{6 \pi \eta r}{\mathrm{v}}\)
3 \(6 \pi \eta r V\)
4 \(\frac{r V}{6 \pi \eta}\)
6 RBTS PAPER

163183 An incompressible liquid flows through a horizontal tube as shown in figure. Then the velocity \(v\) of the fluid is : [A \(\rightarrow\) area of cross section] [RBQ]

1 \(1 \mathrm{~m} / \mathrm{sec}\)
2 \(45 \mathrm{~m} / \mathrm{sec}\)
3 \(5 \mathrm{~m} / \mathrm{sec}\)
4 \(1.5 \mathrm{~m} / \mathrm{sec}\)
6 RBTS PAPER

163184 Water rises to a height of \(\mathbf{1 0} \mathrm{cm}\) in a capillary tube and mercury falles to a depth of \(3.42 \mathrm{~cm}\) in the same capillary tube. If the density of mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^3\) and angle of contact of mercury and water are \(135^{\circ}\) and \(0^{\circ}\) respectively the ratio of surface tension for water and mercury is : [RBQ]

1 \(1: 05\)
2 \(1: 3\)
3 \(1: 6.5\)
4 \(1.5: 1\)
6 RBTS PAPER

163185 The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are \(v_1\) and \(v_2\) respectively. If \(A\) is the cross-sectional area of the wing and ' \(\rho\) ' is the density of air, then the upward lift is: [RBQ]

1 \(\rho \mathrm{A}\left(\mathrm{v}_1-\mathrm{v}_2\right)\)
2 \(\rho \mathrm{A}\left(\mathrm{v}_1+\mathrm{v}_2\right)\)
3 \(\rho \mathrm{A}\left(\mathrm{v}_1^2-\mathrm{v}_2^2\right) / 2\)
4 \(\rho \mathrm{A}\left(\mathrm{v}_1^2+\mathrm{v}_2^2\right)\)
6 RBTS PAPER

163182 The viscous force acting on a body falling under gravity in viscous fluid will be : [RBQ]

1 \(\frac{6 \pi V}{\eta r}\)
2 \(\frac{6 \pi \eta r}{\mathrm{v}}\)
3 \(6 \pi \eta r V\)
4 \(\frac{r V}{6 \pi \eta}\)
6 RBTS PAPER

163183 An incompressible liquid flows through a horizontal tube as shown in figure. Then the velocity \(v\) of the fluid is : [A \(\rightarrow\) area of cross section] [RBQ]

1 \(1 \mathrm{~m} / \mathrm{sec}\)
2 \(45 \mathrm{~m} / \mathrm{sec}\)
3 \(5 \mathrm{~m} / \mathrm{sec}\)
4 \(1.5 \mathrm{~m} / \mathrm{sec}\)
6 RBTS PAPER

163184 Water rises to a height of \(\mathbf{1 0} \mathrm{cm}\) in a capillary tube and mercury falles to a depth of \(3.42 \mathrm{~cm}\) in the same capillary tube. If the density of mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^3\) and angle of contact of mercury and water are \(135^{\circ}\) and \(0^{\circ}\) respectively the ratio of surface tension for water and mercury is : [RBQ]

1 \(1: 05\)
2 \(1: 3\)
3 \(1: 6.5\)
4 \(1.5: 1\)
6 RBTS PAPER

163185 The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are \(v_1\) and \(v_2\) respectively. If \(A\) is the cross-sectional area of the wing and ' \(\rho\) ' is the density of air, then the upward lift is: [RBQ]

1 \(\rho \mathrm{A}\left(\mathrm{v}_1-\mathrm{v}_2\right)\)
2 \(\rho \mathrm{A}\left(\mathrm{v}_1+\mathrm{v}_2\right)\)
3 \(\rho \mathrm{A}\left(\mathrm{v}_1^2-\mathrm{v}_2^2\right) / 2\)
4 \(\rho \mathrm{A}\left(\mathrm{v}_1^2+\mathrm{v}_2^2\right)\)
6 RBTS PAPER

163182 The viscous force acting on a body falling under gravity in viscous fluid will be : [RBQ]

1 \(\frac{6 \pi V}{\eta r}\)
2 \(\frac{6 \pi \eta r}{\mathrm{v}}\)
3 \(6 \pi \eta r V\)
4 \(\frac{r V}{6 \pi \eta}\)
6 RBTS PAPER

163183 An incompressible liquid flows through a horizontal tube as shown in figure. Then the velocity \(v\) of the fluid is : [A \(\rightarrow\) area of cross section] [RBQ]

1 \(1 \mathrm{~m} / \mathrm{sec}\)
2 \(45 \mathrm{~m} / \mathrm{sec}\)
3 \(5 \mathrm{~m} / \mathrm{sec}\)
4 \(1.5 \mathrm{~m} / \mathrm{sec}\)
6 RBTS PAPER

163184 Water rises to a height of \(\mathbf{1 0} \mathrm{cm}\) in a capillary tube and mercury falles to a depth of \(3.42 \mathrm{~cm}\) in the same capillary tube. If the density of mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^3\) and angle of contact of mercury and water are \(135^{\circ}\) and \(0^{\circ}\) respectively the ratio of surface tension for water and mercury is : [RBQ]

1 \(1: 05\)
2 \(1: 3\)
3 \(1: 6.5\)
4 \(1.5: 1\)
6 RBTS PAPER

163185 The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are \(v_1\) and \(v_2\) respectively. If \(A\) is the cross-sectional area of the wing and ' \(\rho\) ' is the density of air, then the upward lift is: [RBQ]

1 \(\rho \mathrm{A}\left(\mathrm{v}_1-\mathrm{v}_2\right)\)
2 \(\rho \mathrm{A}\left(\mathrm{v}_1+\mathrm{v}_2\right)\)
3 \(\rho \mathrm{A}\left(\mathrm{v}_1^2-\mathrm{v}_2^2\right) / 2\)
4 \(\rho \mathrm{A}\left(\mathrm{v}_1^2+\mathrm{v}_2^2\right)\)
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
6 RBTS PAPER

163182 The viscous force acting on a body falling under gravity in viscous fluid will be : [RBQ]

1 \(\frac{6 \pi V}{\eta r}\)
2 \(\frac{6 \pi \eta r}{\mathrm{v}}\)
3 \(6 \pi \eta r V\)
4 \(\frac{r V}{6 \pi \eta}\)
6 RBTS PAPER

163183 An incompressible liquid flows through a horizontal tube as shown in figure. Then the velocity \(v\) of the fluid is : [A \(\rightarrow\) area of cross section] [RBQ]

1 \(1 \mathrm{~m} / \mathrm{sec}\)
2 \(45 \mathrm{~m} / \mathrm{sec}\)
3 \(5 \mathrm{~m} / \mathrm{sec}\)
4 \(1.5 \mathrm{~m} / \mathrm{sec}\)
6 RBTS PAPER

163184 Water rises to a height of \(\mathbf{1 0} \mathrm{cm}\) in a capillary tube and mercury falles to a depth of \(3.42 \mathrm{~cm}\) in the same capillary tube. If the density of mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^3\) and angle of contact of mercury and water are \(135^{\circ}\) and \(0^{\circ}\) respectively the ratio of surface tension for water and mercury is : [RBQ]

1 \(1: 05\)
2 \(1: 3\)
3 \(1: 6.5\)
4 \(1.5: 1\)
6 RBTS PAPER

163185 The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are \(v_1\) and \(v_2\) respectively. If \(A\) is the cross-sectional area of the wing and ' \(\rho\) ' is the density of air, then the upward lift is: [RBQ]

1 \(\rho \mathrm{A}\left(\mathrm{v}_1-\mathrm{v}_2\right)\)
2 \(\rho \mathrm{A}\left(\mathrm{v}_1+\mathrm{v}_2\right)\)
3 \(\rho \mathrm{A}\left(\mathrm{v}_1^2-\mathrm{v}_2^2\right) / 2\)
4 \(\rho \mathrm{A}\left(\mathrm{v}_1^2+\mathrm{v}_2^2\right)\)