142956
Water rises to a height $h$ in a capillary on the surface of the earth in stationary condition. Value of $h$ increases, if this tube is taken-
1 on poles
2 on sun
3 in a lift going upward with acceleration
4 in a lift going downward with acceleration
Explanation:
D We know that, Capillary rise or depression - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ $\Rightarrow \mathrm{h} \propto \frac{1}{\mathrm{~g}}$ When lift going downward with acceleration ' $a$ ' Then, equation of motion $\mathrm{R}+\mathrm{ma}=\mathrm{mg}$ $\mathrm{R}=\mathrm{m}(\mathrm{g}-\mathrm{a})$ i.e. $\mathrm{g}^{\prime}=(\mathrm{g}-\mathrm{a}) \lt \mathrm{g}$ Therefore, $h$ increases.
BCECE-2014
Mechanical Properties of Fluids
142957
Two water pipes of diameters $2 \mathrm{~cm}$ and $4 \mathrm{~cm}$ are connected with the main supply line. If velocity of flow of water in the pipe of $4 \mathrm{~cm}$, diameter is $X$, then velocity in $2 \mathrm{~cm}$, diameter is
1 $2 \mathrm{X}$
2 $4 \mathrm{X}$
3 $6 \mathrm{X}$
4 $8 \mathrm{X}$
Explanation:
B Given, Diameter of pipes, $d_{1}=2 \mathrm{~cm}, d_{2}=4 \mathrm{~cm}$ The velocity along $4 \mathrm{~cm}$ diameter of pipe $\mathrm{v}_{2}=\mathrm{Xm} / \mathrm{s}$ And the velocity along $2 \mathrm{~cm}$ diameter of pipe $=\mathrm{v}_{1}$ From, Continuity equation :- $A_{1} v_{1}=A_{2} v_{2}$ $\frac{\pi}{4}\left(d_{1}\right)^{2} v_{1}=\left(\frac{\pi}{4}\right) d_{2}^{2} v_{2}$ $d_{1}^{2} v_{1}=d_{2}^{2} v_{2}$ $v_{2}=\left(\frac{d_{1}}{d_{2}}\right)^{2} v_{1}$ $X=\left(\frac{2}{4}\right)^{2} v_{1}$ $X=\left(\frac{1}{2}\right)^{2} v_{1}$ $X=\frac{v_{1}}{4}$ $v_{1}=4 X m / s$ Hence the velocity in pipe of $2 \mathrm{~cm}$ diameter is $4 \mathrm{X} \mathrm{m} / \mathrm{s}$.
BCECE-2014
Mechanical Properties of Fluids
142958
When capillary tubes of different radii ' $r$ ' dipped in water, water rises to different heights ' $h$ ' in them, then
1 $h r^{2}=$ constant
2 $\mathrm{hr}=$ constant
3 $\frac{\mathrm{h}}{\mathrm{r}}=$ constant
4 $\frac{\mathrm{h}}{\mathrm{r}^{2}}=$ constant
Explanation:
B We know that, Capillary rise or depression - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ Where, $\mathrm{T}=$ Surface tension $\rho=\text { Density of fluid }$ $\theta=\text { contact angle }$ $r=\text { Radius of capillary tube }$ From equation number (i), $\because$ Tubes of different radii dipped in same liquid (water), $\therefore \theta_{1}=\theta_{2}, \rho_{1}=\rho_{2}, \mathrm{~T}_{1}=\mathrm{T}_{2}$ Capillary rise or depression (h) is inversely proportional to radius of capillary tube. Hence, $\mathrm{h} \propto \frac{1}{\mathrm{r}}$ $\mathrm{hr}=\text { constant }$
BCECE-2013
Mechanical Properties of Fluids
142959
The lower end of a glass capillary tube is dipped in water. Water rises to a height of $9 \mathrm{~cm}$. The tube is then broken at a height of $5 \mathrm{~cm}$. The height of water column and angle of contact will be
A Given, Since, we know that, capillary rise or depression is given as- $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g}$ Initially water is rise at height, $\mathrm{h}_{1}=9 \mathrm{~cm}, \theta_{1}=0^{\circ}$ And if tube is broken, then water rise, From equation (i) $\mathrm{h}_{2}=5 \mathrm{~cm}, \theta_{2}=\text { ? }$ $\mathrm{h} \propto \cos \theta$ $\frac{\mathrm{h}_{1}}{\cos \theta_{1}}=\frac{\mathrm{h}_{2}}{\cos \theta_{2}}$ $\frac{9}{\cos 0^{\circ}}=\frac{5}{\cos \theta_{2}}$ $\cos \theta_{2}=\frac{5}{9}$ $\theta_{2}=\cos ^{-1}(5 / 9)$ So, Height is $5 \mathrm{~cm}$ and contact angle is $\cos ^{-1}(5 / 9)$
142956
Water rises to a height $h$ in a capillary on the surface of the earth in stationary condition. Value of $h$ increases, if this tube is taken-
1 on poles
2 on sun
3 in a lift going upward with acceleration
4 in a lift going downward with acceleration
Explanation:
D We know that, Capillary rise or depression - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ $\Rightarrow \mathrm{h} \propto \frac{1}{\mathrm{~g}}$ When lift going downward with acceleration ' $a$ ' Then, equation of motion $\mathrm{R}+\mathrm{ma}=\mathrm{mg}$ $\mathrm{R}=\mathrm{m}(\mathrm{g}-\mathrm{a})$ i.e. $\mathrm{g}^{\prime}=(\mathrm{g}-\mathrm{a}) \lt \mathrm{g}$ Therefore, $h$ increases.
BCECE-2014
Mechanical Properties of Fluids
142957
Two water pipes of diameters $2 \mathrm{~cm}$ and $4 \mathrm{~cm}$ are connected with the main supply line. If velocity of flow of water in the pipe of $4 \mathrm{~cm}$, diameter is $X$, then velocity in $2 \mathrm{~cm}$, diameter is
1 $2 \mathrm{X}$
2 $4 \mathrm{X}$
3 $6 \mathrm{X}$
4 $8 \mathrm{X}$
Explanation:
B Given, Diameter of pipes, $d_{1}=2 \mathrm{~cm}, d_{2}=4 \mathrm{~cm}$ The velocity along $4 \mathrm{~cm}$ diameter of pipe $\mathrm{v}_{2}=\mathrm{Xm} / \mathrm{s}$ And the velocity along $2 \mathrm{~cm}$ diameter of pipe $=\mathrm{v}_{1}$ From, Continuity equation :- $A_{1} v_{1}=A_{2} v_{2}$ $\frac{\pi}{4}\left(d_{1}\right)^{2} v_{1}=\left(\frac{\pi}{4}\right) d_{2}^{2} v_{2}$ $d_{1}^{2} v_{1}=d_{2}^{2} v_{2}$ $v_{2}=\left(\frac{d_{1}}{d_{2}}\right)^{2} v_{1}$ $X=\left(\frac{2}{4}\right)^{2} v_{1}$ $X=\left(\frac{1}{2}\right)^{2} v_{1}$ $X=\frac{v_{1}}{4}$ $v_{1}=4 X m / s$ Hence the velocity in pipe of $2 \mathrm{~cm}$ diameter is $4 \mathrm{X} \mathrm{m} / \mathrm{s}$.
BCECE-2014
Mechanical Properties of Fluids
142958
When capillary tubes of different radii ' $r$ ' dipped in water, water rises to different heights ' $h$ ' in them, then
1 $h r^{2}=$ constant
2 $\mathrm{hr}=$ constant
3 $\frac{\mathrm{h}}{\mathrm{r}}=$ constant
4 $\frac{\mathrm{h}}{\mathrm{r}^{2}}=$ constant
Explanation:
B We know that, Capillary rise or depression - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ Where, $\mathrm{T}=$ Surface tension $\rho=\text { Density of fluid }$ $\theta=\text { contact angle }$ $r=\text { Radius of capillary tube }$ From equation number (i), $\because$ Tubes of different radii dipped in same liquid (water), $\therefore \theta_{1}=\theta_{2}, \rho_{1}=\rho_{2}, \mathrm{~T}_{1}=\mathrm{T}_{2}$ Capillary rise or depression (h) is inversely proportional to radius of capillary tube. Hence, $\mathrm{h} \propto \frac{1}{\mathrm{r}}$ $\mathrm{hr}=\text { constant }$
BCECE-2013
Mechanical Properties of Fluids
142959
The lower end of a glass capillary tube is dipped in water. Water rises to a height of $9 \mathrm{~cm}$. The tube is then broken at a height of $5 \mathrm{~cm}$. The height of water column and angle of contact will be
A Given, Since, we know that, capillary rise or depression is given as- $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g}$ Initially water is rise at height, $\mathrm{h}_{1}=9 \mathrm{~cm}, \theta_{1}=0^{\circ}$ And if tube is broken, then water rise, From equation (i) $\mathrm{h}_{2}=5 \mathrm{~cm}, \theta_{2}=\text { ? }$ $\mathrm{h} \propto \cos \theta$ $\frac{\mathrm{h}_{1}}{\cos \theta_{1}}=\frac{\mathrm{h}_{2}}{\cos \theta_{2}}$ $\frac{9}{\cos 0^{\circ}}=\frac{5}{\cos \theta_{2}}$ $\cos \theta_{2}=\frac{5}{9}$ $\theta_{2}=\cos ^{-1}(5 / 9)$ So, Height is $5 \mathrm{~cm}$ and contact angle is $\cos ^{-1}(5 / 9)$
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Mechanical Properties of Fluids
142956
Water rises to a height $h$ in a capillary on the surface of the earth in stationary condition. Value of $h$ increases, if this tube is taken-
1 on poles
2 on sun
3 in a lift going upward with acceleration
4 in a lift going downward with acceleration
Explanation:
D We know that, Capillary rise or depression - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ $\Rightarrow \mathrm{h} \propto \frac{1}{\mathrm{~g}}$ When lift going downward with acceleration ' $a$ ' Then, equation of motion $\mathrm{R}+\mathrm{ma}=\mathrm{mg}$ $\mathrm{R}=\mathrm{m}(\mathrm{g}-\mathrm{a})$ i.e. $\mathrm{g}^{\prime}=(\mathrm{g}-\mathrm{a}) \lt \mathrm{g}$ Therefore, $h$ increases.
BCECE-2014
Mechanical Properties of Fluids
142957
Two water pipes of diameters $2 \mathrm{~cm}$ and $4 \mathrm{~cm}$ are connected with the main supply line. If velocity of flow of water in the pipe of $4 \mathrm{~cm}$, diameter is $X$, then velocity in $2 \mathrm{~cm}$, diameter is
1 $2 \mathrm{X}$
2 $4 \mathrm{X}$
3 $6 \mathrm{X}$
4 $8 \mathrm{X}$
Explanation:
B Given, Diameter of pipes, $d_{1}=2 \mathrm{~cm}, d_{2}=4 \mathrm{~cm}$ The velocity along $4 \mathrm{~cm}$ diameter of pipe $\mathrm{v}_{2}=\mathrm{Xm} / \mathrm{s}$ And the velocity along $2 \mathrm{~cm}$ diameter of pipe $=\mathrm{v}_{1}$ From, Continuity equation :- $A_{1} v_{1}=A_{2} v_{2}$ $\frac{\pi}{4}\left(d_{1}\right)^{2} v_{1}=\left(\frac{\pi}{4}\right) d_{2}^{2} v_{2}$ $d_{1}^{2} v_{1}=d_{2}^{2} v_{2}$ $v_{2}=\left(\frac{d_{1}}{d_{2}}\right)^{2} v_{1}$ $X=\left(\frac{2}{4}\right)^{2} v_{1}$ $X=\left(\frac{1}{2}\right)^{2} v_{1}$ $X=\frac{v_{1}}{4}$ $v_{1}=4 X m / s$ Hence the velocity in pipe of $2 \mathrm{~cm}$ diameter is $4 \mathrm{X} \mathrm{m} / \mathrm{s}$.
BCECE-2014
Mechanical Properties of Fluids
142958
When capillary tubes of different radii ' $r$ ' dipped in water, water rises to different heights ' $h$ ' in them, then
1 $h r^{2}=$ constant
2 $\mathrm{hr}=$ constant
3 $\frac{\mathrm{h}}{\mathrm{r}}=$ constant
4 $\frac{\mathrm{h}}{\mathrm{r}^{2}}=$ constant
Explanation:
B We know that, Capillary rise or depression - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ Where, $\mathrm{T}=$ Surface tension $\rho=\text { Density of fluid }$ $\theta=\text { contact angle }$ $r=\text { Radius of capillary tube }$ From equation number (i), $\because$ Tubes of different radii dipped in same liquid (water), $\therefore \theta_{1}=\theta_{2}, \rho_{1}=\rho_{2}, \mathrm{~T}_{1}=\mathrm{T}_{2}$ Capillary rise or depression (h) is inversely proportional to radius of capillary tube. Hence, $\mathrm{h} \propto \frac{1}{\mathrm{r}}$ $\mathrm{hr}=\text { constant }$
BCECE-2013
Mechanical Properties of Fluids
142959
The lower end of a glass capillary tube is dipped in water. Water rises to a height of $9 \mathrm{~cm}$. The tube is then broken at a height of $5 \mathrm{~cm}$. The height of water column and angle of contact will be
A Given, Since, we know that, capillary rise or depression is given as- $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g}$ Initially water is rise at height, $\mathrm{h}_{1}=9 \mathrm{~cm}, \theta_{1}=0^{\circ}$ And if tube is broken, then water rise, From equation (i) $\mathrm{h}_{2}=5 \mathrm{~cm}, \theta_{2}=\text { ? }$ $\mathrm{h} \propto \cos \theta$ $\frac{\mathrm{h}_{1}}{\cos \theta_{1}}=\frac{\mathrm{h}_{2}}{\cos \theta_{2}}$ $\frac{9}{\cos 0^{\circ}}=\frac{5}{\cos \theta_{2}}$ $\cos \theta_{2}=\frac{5}{9}$ $\theta_{2}=\cos ^{-1}(5 / 9)$ So, Height is $5 \mathrm{~cm}$ and contact angle is $\cos ^{-1}(5 / 9)$
142956
Water rises to a height $h$ in a capillary on the surface of the earth in stationary condition. Value of $h$ increases, if this tube is taken-
1 on poles
2 on sun
3 in a lift going upward with acceleration
4 in a lift going downward with acceleration
Explanation:
D We know that, Capillary rise or depression - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ $\Rightarrow \mathrm{h} \propto \frac{1}{\mathrm{~g}}$ When lift going downward with acceleration ' $a$ ' Then, equation of motion $\mathrm{R}+\mathrm{ma}=\mathrm{mg}$ $\mathrm{R}=\mathrm{m}(\mathrm{g}-\mathrm{a})$ i.e. $\mathrm{g}^{\prime}=(\mathrm{g}-\mathrm{a}) \lt \mathrm{g}$ Therefore, $h$ increases.
BCECE-2014
Mechanical Properties of Fluids
142957
Two water pipes of diameters $2 \mathrm{~cm}$ and $4 \mathrm{~cm}$ are connected with the main supply line. If velocity of flow of water in the pipe of $4 \mathrm{~cm}$, diameter is $X$, then velocity in $2 \mathrm{~cm}$, diameter is
1 $2 \mathrm{X}$
2 $4 \mathrm{X}$
3 $6 \mathrm{X}$
4 $8 \mathrm{X}$
Explanation:
B Given, Diameter of pipes, $d_{1}=2 \mathrm{~cm}, d_{2}=4 \mathrm{~cm}$ The velocity along $4 \mathrm{~cm}$ diameter of pipe $\mathrm{v}_{2}=\mathrm{Xm} / \mathrm{s}$ And the velocity along $2 \mathrm{~cm}$ diameter of pipe $=\mathrm{v}_{1}$ From, Continuity equation :- $A_{1} v_{1}=A_{2} v_{2}$ $\frac{\pi}{4}\left(d_{1}\right)^{2} v_{1}=\left(\frac{\pi}{4}\right) d_{2}^{2} v_{2}$ $d_{1}^{2} v_{1}=d_{2}^{2} v_{2}$ $v_{2}=\left(\frac{d_{1}}{d_{2}}\right)^{2} v_{1}$ $X=\left(\frac{2}{4}\right)^{2} v_{1}$ $X=\left(\frac{1}{2}\right)^{2} v_{1}$ $X=\frac{v_{1}}{4}$ $v_{1}=4 X m / s$ Hence the velocity in pipe of $2 \mathrm{~cm}$ diameter is $4 \mathrm{X} \mathrm{m} / \mathrm{s}$.
BCECE-2014
Mechanical Properties of Fluids
142958
When capillary tubes of different radii ' $r$ ' dipped in water, water rises to different heights ' $h$ ' in them, then
1 $h r^{2}=$ constant
2 $\mathrm{hr}=$ constant
3 $\frac{\mathrm{h}}{\mathrm{r}}=$ constant
4 $\frac{\mathrm{h}}{\mathrm{r}^{2}}=$ constant
Explanation:
B We know that, Capillary rise or depression - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ Where, $\mathrm{T}=$ Surface tension $\rho=\text { Density of fluid }$ $\theta=\text { contact angle }$ $r=\text { Radius of capillary tube }$ From equation number (i), $\because$ Tubes of different radii dipped in same liquid (water), $\therefore \theta_{1}=\theta_{2}, \rho_{1}=\rho_{2}, \mathrm{~T}_{1}=\mathrm{T}_{2}$ Capillary rise or depression (h) is inversely proportional to radius of capillary tube. Hence, $\mathrm{h} \propto \frac{1}{\mathrm{r}}$ $\mathrm{hr}=\text { constant }$
BCECE-2013
Mechanical Properties of Fluids
142959
The lower end of a glass capillary tube is dipped in water. Water rises to a height of $9 \mathrm{~cm}$. The tube is then broken at a height of $5 \mathrm{~cm}$. The height of water column and angle of contact will be
A Given, Since, we know that, capillary rise or depression is given as- $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g}$ Initially water is rise at height, $\mathrm{h}_{1}=9 \mathrm{~cm}, \theta_{1}=0^{\circ}$ And if tube is broken, then water rise, From equation (i) $\mathrm{h}_{2}=5 \mathrm{~cm}, \theta_{2}=\text { ? }$ $\mathrm{h} \propto \cos \theta$ $\frac{\mathrm{h}_{1}}{\cos \theta_{1}}=\frac{\mathrm{h}_{2}}{\cos \theta_{2}}$ $\frac{9}{\cos 0^{\circ}}=\frac{5}{\cos \theta_{2}}$ $\cos \theta_{2}=\frac{5}{9}$ $\theta_{2}=\cos ^{-1}(5 / 9)$ So, Height is $5 \mathrm{~cm}$ and contact angle is $\cos ^{-1}(5 / 9)$