JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 19)
As shown in figure, when a spherical cavity
(centered at O) of radius 1 is cut out of a uniform
sphere of radius R (centered at C), the centre of
mass of remaining (shaded) part of sphere is at
G, i.e, on the surface of the cavity. R can be
detemined by the equation :
_8th_January_Evening_Slot_en_19_1.png)
(R2 + R – 1) (2 – R) = 1
(R2 – R – 1) (2 – R) = 1
(R2 – R + 1) (2 – R) = 1
(R2 + R + 1) (2 – R) = 1
Explanation
_8th_January_Evening_Slot_en_19_2.png)
By concept of COM
m0R0 = mcavityRcavity
$$ \Rightarrow $$ $$\left( {{4 \over 3}\pi {R^3}\rho - {4 \over 3}\pi {{\left( 1 \right)}^3}\rho } \right)$$ (2 - R) = $${{4 \over 3}\pi {{\left( 1 \right)}^3}\rho }$$ $$ \times $$ (R - 1)
$$ \Rightarrow $$ (R3 – 1) (2 – R) = R – 1
$$ \Rightarrow $$(R2 + R + 1) (2 – R) = 1
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