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Question

$(a)$ Given, drag force $F_{d}$ depends on air density $\sigma$, velocity of ball $v$ and area of cross-section of ball $A$.

$	herefore \quad F_{d

Vedclass · Accepted Answer

$2^{1 / 6}$

Vedclass · Answer

$(a)$ Given, drag force $F_{d}$ depends on air density $\sigma$, velocity of ball $v$ and area of cross-section of ball $A$.

$	herefore \quad F_{d} \propto \sigma^{\alpha}\propto v^{b} A^{c}$

Substituting dimensions of different physical quantities, we have

$\left[ MLT ^{-2}ight]=\left[ ML ^{-3}ight]^{\alpha}\left[ LT ^{-1}ight]^{b}\left[ L ^{2}ight]^{c}$

Equating dimensions of fundamental quantities, we have

$a=1 \quad \dots(i)$

$-3 a+b+2 c=1\quad \dots(ii)$

$-b=-2 \Rightarrow b=2\quad \dots(iii)$

Putting the values of $a$ and $b$ in Eq $(ii)$, we get

$\Rightarrow \quad-3 	imes 1+2+2 c=1 \Rightarrow c=1$

$\because a=1, b=2, c=1$

So, drag force on football is

$F_{d}=k \cdot \sigma \cdot v^{2} \cdot A$

When football reaches terminal speed, its weight is balanced by drag force.

$\Rightarrow m g=F_{d}$

$\Rightarrow m g=k \sigma v_{T}^{2} A$

where, $u_{T}=$ terminal speed of football.

$\Rightarrow \quad m g=k \sigma v_{T}^{2} \pi R^{2}$

$\Rightarrow \quad m g=k \sigma u_{T}^{2} \pi\left(\frac{m}{\frac{4}{3} \pi ho}ight)^{\frac{2}{3}}$

[from $m=\frac{4}{3} \pi R^{3} ho$,

where $ho=$ density of football]

$\Rightarrow \quad u_{T} \propto m^{1 / 6}$

$\Rightarrow \quad \frac{v_{1}}{v_{2}}=\left(\frac{m_{1}}{m_{2}}ight)^{\frac{1}{6}}=\left(\frac{250}{125}ight)^{\frac{1}{6}}=2^{1 / 6}$

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