CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A solid sphere of uniform density and radius $$R$$ applies a gravitational force of attraction equal to $$F_1$$ on a particle placed at a distance 3R from the centre of the sphere. A spherical cavity of radius $$\dfrac{R}{2}$$ is now made in the sphere, as shown in the figure. The sphere with cavity now applies a gravitational force $$F_2$$ on the same particle. The ratio $$\dfrac{F_2}{F_1}$$ is
6162.png


A
950
loader
B
4150
loader
C
325
loader
D
2225
loader

Solution

The correct option is A $$\dfrac{41}{50}$$

From super position principle , 

$$F_{1}=F_{r}+F_{c}$$

Here $$F_{r}=$$ force due to remaining part $$=F_{2} $$

$$F_{c}=$$ force due to mass on the cavity

$$F_{1}=\dfrac{GMm}{9R^{2}}$$

$$\displaystyle {F_{c}=\dfrac{G(\dfrac{M}{8})m}{(\dfrac{5}{2}R)^{2}}=\dfrac{GMm}{50R^{2}}}$$

$$=\displaystyle \succ F_{2}=F_{1}-F_{c}=\dfrac{GMm}{9R^{2}}-\dfrac{GMm}{50R^{2}}=\dfrac{41GMm}{450R^{2}}$$

$$\Rightarrow  \displaystyle \dfrac{F_{2}}{F_{1}}=\dfrac{41}{50}$$


Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image