Given: The ratio of masses of two planets \(M_{1}: M_{2} \) is 2: 3 The ratio of their radii \(R_{1}: R_{2}\) is 4: 7
\(g_{1}: g_{2}= ? \)
\(\mathrm{g}=\frac{\mathrm{G \times M}}{\mathrm{R}^{2}} ; \quad \mathrm{g}_{1}=\frac{\mathrm{G\times M}_{1}}{\mathrm{R}_{1}^{2}} ; \quad \mathrm{g}_{2}=\frac{\mathrm{G \times M}_{2}}{\mathrm{R}_{2}^{2}} \)
\(\frac{g_{1}}{g_{2}}=\frac{\frac{G \times M_{1}}{R_{1}^{2}}}{\frac{G \times M_{2}}{R_{2}^{2}}} ; \quad \frac{g_{1}}{g_{2}}=\frac{M_{1}}{R_{1}^{2}} \times \frac{R_{2}^{2}}{M_{2}}=\frac{2 \times 7 \times 7}{4 \times 4 \times 3}=\frac{49}{24}\)
∴ \(g_{1}: g_{2}\) = 49: 24

\(\text { Given } \quad \mathrm{F}_{1} =140 \mathrm{~N}  \)
\(\mathrm{~d}_{1} =40 \mathrm{~cm}=40 \times 10^{-2} \mathrm{~m}  \)
\(\mathrm{~F}_{2} =40 \mathrm{~N}  \)
\(\mathrm{~d}_{2} =?\)
The moment of force on the nut,
\(=F_{1} \times d_{1}=140 \times 40 \times 10^{-2}  \)
\(=56 \ \mathrm{Nm}\)
The same moment of force is required to unscrew the nut,
\(\mathrm{F}_{2} \times \mathrm{~d}_{2}  =40 \times d_2\)
In both cases, moment of forces applied are equal,
F₁ x d₁ = F₂ x d₂
56 = F₂d₂
\(56=40 \times \mathrm{d}_{2}  \)
\(\mathrm{~d}_{2}=\frac{56}{40}=1.4 \mathrm{~m}  \)
\(\mathrm{~d}_{2}=1.4 \mathrm{~m}\)
The spanner with the longer handle requires very less force to unscrew the same nut.

Given \(\mathrm{m}=1 \mathrm{~kg}, \quad \mathrm{v}=10 \mathrm{~m} \mathrm{~s}^{-1}\)
When a ball bounces back with the same speed, the momentum changes from mv to -mv. So, the change in momentum is -2 mv.
Δp = mv - mu
= -mv - mv
\(=-2 \mathrm{mv}=-2 \times 1 \times 10 \)      \([Here \ mu = mv \\ mv = -mv]\)
\(\therefore \Delta P=-20 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)

\(\text {Given } m_{1} : m_{2}=3: 4  \)          
\(a_{2} =12 \mathrm{~m} \mathrm{~s}^{-2}  \)
\(a_{1} =?\)
Using Newton's second law,
\(\mathrm{F} =\mathrm{m \times a}  \)
\(\mathrm{F} =\mathrm{m}_{1} \times a_{1}=3 a_{1}  \)
\(\mathrm{~F} =\mathrm{m}_{2} \times a_{2}=4 \times 12=48 \mathrm{~N}  \)
\(3 a_{1} =48  \)
\(a_{1} =48 / 3=16 \mathrm{~m} \mathrm{~s}^{-2}  \)
\(a_{1} =16 \mathrm{~ms}^{-2}\)
The acceleration produced on the other body is \(16 \mathrm{~m} \mathrm{~s}^{-2}\).