Question

Any JEE Aspirant or physics lover.

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Answer 1

Explanation:

its simple dude

v^2-u^2=2as

(30)^2-0=2×a×10^-3

a=4.5×10^5

f=ma

=(4×10^-3-3)(4.5×10^5)

=18×10^-1

=1.8

hope it helps you dude

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Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

Raindrops of radius 1 mm and mass 4 mg are falling with a speed of 30 m/s on head of a bald person. The drops splash on the head and come to rest. assuming equivalently that the drops cover a distance equal to their on the head,estimate the force excerted by each drop on head?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of the drop (M) = 4mg = 4 × 10⁻⁶Kg.
• Radius of Drop (r) = 1mm = 10⁻³ m.
• Final velocity (v) = 0 m/s.
• Initial velocity (u) = 30 m/s.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Now, When the Drop falls on the bald head,

The distance it travels equals The radius of the Drop.

∴ S = r = 10⁻³ m.

Applying Third kinematic equation,

$\large{\boxed{\tt v^2 - u^2 = 2as}}$

Substituting the values,

$\large{\tt \leadsto (0)^2 - (30)^2 = 2 \times a \times 10^{-3}}$

• Here Final velocity will be Zero, as It comes to rest after colliding with the head.

Simplifying,

$\large{\tt \leadsto 0 - 900 = 2 \times 10^{-3} \times a}$

$\large{\tt \leadsto - 900 = 2 \times 10^{-3} \times a}$

$\large{\tt \leadsto a = \dfrac{- 900}{2 \times 10^{-3}}}$

$\large{\tt \leadsto a = \dfrac{\cancel{- 900}}{\cancel{2} \times 10^{-3}}}$

$\large{\tt \leadsto a = \dfrac{- 450}{10^{-3}}}$

$\large{\tt \leadsto a = - 450 \times 10^3}$

$\large{\tt \leadsto a = - 45 \times 10^4}$

Taking Magnitude,

• Here Negative sign indicates that the body is retarding.

$\large{\boxed{\tt a = 45 \times 10^4 \: m/s^2}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Applying Newton's second law of motion,

$\large{\boxed{\tt F = Ma}}$

Substituting the values,

$\large{\tt \leadsto F = 4mg \times 45 \times 10^4}$

$\large{\tt \leadsto F = 4 \times 10^{-6} \times 45 \times 10^4 \: \longrightarrow 4mg = 4 \times 10^{-6} \: Kg}$

Simplifying,

$\large{\tt \leadsto F = 4 \times 45 \times 10^{-6 + 4}}$

$\large{\tt \leadsto F = 180 \times 10^{-2}}$

Now,

$\huge{\boxed{\boxed{\tt F = 1.8 \: N}}}$

So, the Force Exerted by the Drop on the head is 1.8 Newton.

$\rule{300}{1.5}$