142713
A copper ball of radius $r$ is moving with a uniform velocity $v$ in the mustard oil and the dragging force acting on the ball is $F$. The dragging force on the copper ball of radius $2 r$ with uniform velocity $2 \mathrm{v}$ in the mustard oil is
1 $\mathrm{F}$
2 $2 \mathrm{~F}$
3 $4 \mathrm{~F}$
4 $8 \mathrm{~F}$
Explanation:
C Given, Radius of copper ball $=\mathrm{r}$ Velocity of ball $=\mathrm{v}$ According to Stoke's law drag force or viscous force is given as, Ans: c : Given that, Radius of Sphere $=\mathrm{a}$, Velocity $=\mathrm{v}$, Viscosity of liquid $=\eta$ So, the dimension method opposing force depends upon $F \propto \eta^{a} a^{b} v^{c}$ Writing dimension both side $\mathrm{F}=\mathrm{k \eta} \eta^{\mathrm{a}} \mathrm{a}^{\mathrm{b}} \mathrm{v}^{\mathrm{c}} \text { (A) }$ ${\left[\mathrm{MLT}^{-2}\right]=\mathrm{k}\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]^{\mathrm{a}}[\mathrm{L}]^{\mathrm{b}}\left[\mathrm{LT}^{-1}\right]}$ On comparing both side $a=1$ $-a+b+c=1$ $-a-c=-2$ On solving equation (I) (II) and (III) $\mathrm{a}=1 \quad \mathrm{~b}=1 \quad \mathrm{c}=1$ Putting these value in equation (A) $\mathrm{F}=\mathrm{k \eta av}$ $\mathrm{F}=6 \pi \eta \mathrm{av}$ Where $\mathrm{k}$ is a constant and $\mathrm{k}=6 \pi$
From equation (i) and (ii)
Mechanical Properties of Fluids
142714
Two solid spheres of same metal but of mass $M$ and $8 \mathrm{M}$ fall simultaneously on a viscous liquid and their terminal velocities are $v$ and $n v$, then value of $\mathbf{n}$ is
1 16
2 8
3 4
4 2
Explanation:
C The terminal velocity is given by $\mathrm{v}$. where $v \propto r^{2}(r=$ Radius of the sphere $)$ The mass of the sphere can be given by mass $=$ volume $\times$ density $\therefore \quad \mathrm{m}=\frac{4}{3} \pi \mathrm{r}^{3} \times \rho$ $\Rightarrow \quad \mathrm{m} \propto \mathrm{r}^{3}$ or $\mathrm{r} \propto \mathrm{m}^{1 / 3}$ $\therefore \quad \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)^{1 / 3}=\frac{1}{2}$ And since, $\begin{array}{ll} \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2} \\ \Rightarrow \quad \frac{\mathrm{v}}{\mathrm{nv}}=\left(\frac{1}{2}\right)^{2} \\ \Rightarrow \quad \frac{1}{\mathrm{n}}=\frac{1}{4}\\ \Rightarrow \quad \mathrm{n}=4 \end{array}$
WB JEE 2010
Mechanical Properties of Fluids
142716
When the room temperature becomes equal to the dew point, the relative humidity of the room is
1 $100 \%$
2 zero \%
3 $70 \%$
4 $85 \%$
Explanation:
A We know that, Relative humidity $=$ $=\frac{\text { saturation vapour pressure at dew point }}{\text { sat.vapour.pressure at room temperature }} \times 100 \%$ Given in question, $\text { room temperature }=\text { dew point. }$ So, Relative humidity $=1 \times 100=100 \%$
142713
A copper ball of radius $r$ is moving with a uniform velocity $v$ in the mustard oil and the dragging force acting on the ball is $F$. The dragging force on the copper ball of radius $2 r$ with uniform velocity $2 \mathrm{v}$ in the mustard oil is
1 $\mathrm{F}$
2 $2 \mathrm{~F}$
3 $4 \mathrm{~F}$
4 $8 \mathrm{~F}$
Explanation:
C Given, Radius of copper ball $=\mathrm{r}$ Velocity of ball $=\mathrm{v}$ According to Stoke's law drag force or viscous force is given as, Ans: c : Given that, Radius of Sphere $=\mathrm{a}$, Velocity $=\mathrm{v}$, Viscosity of liquid $=\eta$ So, the dimension method opposing force depends upon $F \propto \eta^{a} a^{b} v^{c}$ Writing dimension both side $\mathrm{F}=\mathrm{k \eta} \eta^{\mathrm{a}} \mathrm{a}^{\mathrm{b}} \mathrm{v}^{\mathrm{c}} \text { (A) }$ ${\left[\mathrm{MLT}^{-2}\right]=\mathrm{k}\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]^{\mathrm{a}}[\mathrm{L}]^{\mathrm{b}}\left[\mathrm{LT}^{-1}\right]}$ On comparing both side $a=1$ $-a+b+c=1$ $-a-c=-2$ On solving equation (I) (II) and (III) $\mathrm{a}=1 \quad \mathrm{~b}=1 \quad \mathrm{c}=1$ Putting these value in equation (A) $\mathrm{F}=\mathrm{k \eta av}$ $\mathrm{F}=6 \pi \eta \mathrm{av}$ Where $\mathrm{k}$ is a constant and $\mathrm{k}=6 \pi$
From equation (i) and (ii)
Mechanical Properties of Fluids
142714
Two solid spheres of same metal but of mass $M$ and $8 \mathrm{M}$ fall simultaneously on a viscous liquid and their terminal velocities are $v$ and $n v$, then value of $\mathbf{n}$ is
1 16
2 8
3 4
4 2
Explanation:
C The terminal velocity is given by $\mathrm{v}$. where $v \propto r^{2}(r=$ Radius of the sphere $)$ The mass of the sphere can be given by mass $=$ volume $\times$ density $\therefore \quad \mathrm{m}=\frac{4}{3} \pi \mathrm{r}^{3} \times \rho$ $\Rightarrow \quad \mathrm{m} \propto \mathrm{r}^{3}$ or $\mathrm{r} \propto \mathrm{m}^{1 / 3}$ $\therefore \quad \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)^{1 / 3}=\frac{1}{2}$ And since, $\begin{array}{ll} \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2} \\ \Rightarrow \quad \frac{\mathrm{v}}{\mathrm{nv}}=\left(\frac{1}{2}\right)^{2} \\ \Rightarrow \quad \frac{1}{\mathrm{n}}=\frac{1}{4}\\ \Rightarrow \quad \mathrm{n}=4 \end{array}$
WB JEE 2010
Mechanical Properties of Fluids
142716
When the room temperature becomes equal to the dew point, the relative humidity of the room is
1 $100 \%$
2 zero \%
3 $70 \%$
4 $85 \%$
Explanation:
A We know that, Relative humidity $=$ $=\frac{\text { saturation vapour pressure at dew point }}{\text { sat.vapour.pressure at room temperature }} \times 100 \%$ Given in question, $\text { room temperature }=\text { dew point. }$ So, Relative humidity $=1 \times 100=100 \%$
142713
A copper ball of radius $r$ is moving with a uniform velocity $v$ in the mustard oil and the dragging force acting on the ball is $F$. The dragging force on the copper ball of radius $2 r$ with uniform velocity $2 \mathrm{v}$ in the mustard oil is
1 $\mathrm{F}$
2 $2 \mathrm{~F}$
3 $4 \mathrm{~F}$
4 $8 \mathrm{~F}$
Explanation:
C Given, Radius of copper ball $=\mathrm{r}$ Velocity of ball $=\mathrm{v}$ According to Stoke's law drag force or viscous force is given as, Ans: c : Given that, Radius of Sphere $=\mathrm{a}$, Velocity $=\mathrm{v}$, Viscosity of liquid $=\eta$ So, the dimension method opposing force depends upon $F \propto \eta^{a} a^{b} v^{c}$ Writing dimension both side $\mathrm{F}=\mathrm{k \eta} \eta^{\mathrm{a}} \mathrm{a}^{\mathrm{b}} \mathrm{v}^{\mathrm{c}} \text { (A) }$ ${\left[\mathrm{MLT}^{-2}\right]=\mathrm{k}\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]^{\mathrm{a}}[\mathrm{L}]^{\mathrm{b}}\left[\mathrm{LT}^{-1}\right]}$ On comparing both side $a=1$ $-a+b+c=1$ $-a-c=-2$ On solving equation (I) (II) and (III) $\mathrm{a}=1 \quad \mathrm{~b}=1 \quad \mathrm{c}=1$ Putting these value in equation (A) $\mathrm{F}=\mathrm{k \eta av}$ $\mathrm{F}=6 \pi \eta \mathrm{av}$ Where $\mathrm{k}$ is a constant and $\mathrm{k}=6 \pi$
From equation (i) and (ii)
Mechanical Properties of Fluids
142714
Two solid spheres of same metal but of mass $M$ and $8 \mathrm{M}$ fall simultaneously on a viscous liquid and their terminal velocities are $v$ and $n v$, then value of $\mathbf{n}$ is
1 16
2 8
3 4
4 2
Explanation:
C The terminal velocity is given by $\mathrm{v}$. where $v \propto r^{2}(r=$ Radius of the sphere $)$ The mass of the sphere can be given by mass $=$ volume $\times$ density $\therefore \quad \mathrm{m}=\frac{4}{3} \pi \mathrm{r}^{3} \times \rho$ $\Rightarrow \quad \mathrm{m} \propto \mathrm{r}^{3}$ or $\mathrm{r} \propto \mathrm{m}^{1 / 3}$ $\therefore \quad \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)^{1 / 3}=\frac{1}{2}$ And since, $\begin{array}{ll} \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2} \\ \Rightarrow \quad \frac{\mathrm{v}}{\mathrm{nv}}=\left(\frac{1}{2}\right)^{2} \\ \Rightarrow \quad \frac{1}{\mathrm{n}}=\frac{1}{4}\\ \Rightarrow \quad \mathrm{n}=4 \end{array}$
WB JEE 2010
Mechanical Properties of Fluids
142716
When the room temperature becomes equal to the dew point, the relative humidity of the room is
1 $100 \%$
2 zero \%
3 $70 \%$
4 $85 \%$
Explanation:
A We know that, Relative humidity $=$ $=\frac{\text { saturation vapour pressure at dew point }}{\text { sat.vapour.pressure at room temperature }} \times 100 \%$ Given in question, $\text { room temperature }=\text { dew point. }$ So, Relative humidity $=1 \times 100=100 \%$
142713
A copper ball of radius $r$ is moving with a uniform velocity $v$ in the mustard oil and the dragging force acting on the ball is $F$. The dragging force on the copper ball of radius $2 r$ with uniform velocity $2 \mathrm{v}$ in the mustard oil is
1 $\mathrm{F}$
2 $2 \mathrm{~F}$
3 $4 \mathrm{~F}$
4 $8 \mathrm{~F}$
Explanation:
C Given, Radius of copper ball $=\mathrm{r}$ Velocity of ball $=\mathrm{v}$ According to Stoke's law drag force or viscous force is given as, Ans: c : Given that, Radius of Sphere $=\mathrm{a}$, Velocity $=\mathrm{v}$, Viscosity of liquid $=\eta$ So, the dimension method opposing force depends upon $F \propto \eta^{a} a^{b} v^{c}$ Writing dimension both side $\mathrm{F}=\mathrm{k \eta} \eta^{\mathrm{a}} \mathrm{a}^{\mathrm{b}} \mathrm{v}^{\mathrm{c}} \text { (A) }$ ${\left[\mathrm{MLT}^{-2}\right]=\mathrm{k}\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]^{\mathrm{a}}[\mathrm{L}]^{\mathrm{b}}\left[\mathrm{LT}^{-1}\right]}$ On comparing both side $a=1$ $-a+b+c=1$ $-a-c=-2$ On solving equation (I) (II) and (III) $\mathrm{a}=1 \quad \mathrm{~b}=1 \quad \mathrm{c}=1$ Putting these value in equation (A) $\mathrm{F}=\mathrm{k \eta av}$ $\mathrm{F}=6 \pi \eta \mathrm{av}$ Where $\mathrm{k}$ is a constant and $\mathrm{k}=6 \pi$
From equation (i) and (ii)
Mechanical Properties of Fluids
142714
Two solid spheres of same metal but of mass $M$ and $8 \mathrm{M}$ fall simultaneously on a viscous liquid and their terminal velocities are $v$ and $n v$, then value of $\mathbf{n}$ is
1 16
2 8
3 4
4 2
Explanation:
C The terminal velocity is given by $\mathrm{v}$. where $v \propto r^{2}(r=$ Radius of the sphere $)$ The mass of the sphere can be given by mass $=$ volume $\times$ density $\therefore \quad \mathrm{m}=\frac{4}{3} \pi \mathrm{r}^{3} \times \rho$ $\Rightarrow \quad \mathrm{m} \propto \mathrm{r}^{3}$ or $\mathrm{r} \propto \mathrm{m}^{1 / 3}$ $\therefore \quad \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)^{1 / 3}=\frac{1}{2}$ And since, $\begin{array}{ll} \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2} \\ \Rightarrow \quad \frac{\mathrm{v}}{\mathrm{nv}}=\left(\frac{1}{2}\right)^{2} \\ \Rightarrow \quad \frac{1}{\mathrm{n}}=\frac{1}{4}\\ \Rightarrow \quad \mathrm{n}=4 \end{array}$
WB JEE 2010
Mechanical Properties of Fluids
142716
When the room temperature becomes equal to the dew point, the relative humidity of the room is
1 $100 \%$
2 zero \%
3 $70 \%$
4 $85 \%$
Explanation:
A We know that, Relative humidity $=$ $=\frac{\text { saturation vapour pressure at dew point }}{\text { sat.vapour.pressure at room temperature }} \times 100 \%$ Given in question, $\text { room temperature }=\text { dew point. }$ So, Relative humidity $=1 \times 100=100 \%$