A uniform solid cone of height H is cut at a distance × below its vertex, parallel to its base and the upper part is discarded. The C.M. of the remaining part is found to be 50 % lower, w.r.t. to the base. The ratio, r=(x/H) satisfies.

Question

A uniform solid cone of height H is cut at a distance × below its vertex, parallel to its base and the upper part is discarded. The C.M. of the remaining part is found to be 50 % lower, w.r.t. to the base. The ratio, r=(x/H) satisfies. (A) 7r3-8r4=1 (B) r=(1/21 / 3) (C) r=(1/2) (D) 7r3-6r4=1

Tardigrade Admin · Accepted Answer

The final C.M. is at (H/8) above the base. The volume, and consequently the mass of a cone (of fixed vertical angle) is proportional to the cube of its height. Taking the origin at apex. (7 H/8)=(H3 ((3 H/4)) - x3 ((3 x/4))/H3 - x3) Rearranging, we get the required condition. ⇒ (7 H/8)× (4/3)=(H4 - x4/H3 - x3) ⇒ (7 H/6)=(H4 - x4/H3 - x3) ⇒ (7/6)=(1 - r4/1 - r3) ⇒ 7(1 - r3)=6(1 - r4) ⇒ 7-7r3=6-6r4 ⇒ 7r3-6r4=1

Q. A uniform solid cone of height $H$ is cut at a distance $\times$ below its vertex, parallel to its base and the upper part is discarded. The $C . M .$ of the remaining part is found to be $50%$ lower, w.r.t. to the base. The ratio, $r = \frac{x}{H}$ satisfies.

$7 r^{3} - 8 r^{4} = 1$

$r = \frac{1}{2 ^{1/3}}$

$r = \frac{1}{2}$

$7 r^{3} - 6 r^{4} = 1$

Q. A uniform solid cone of height H is cut at a distance × below its vertex, parallel to its base and the upper part is discarded. The C.M. of the remaining part is found to be 50% lower, w.r.t. to the base. The ratio, r=Hx​ satisfies.

7r3−8r4=1

r=21/31​

r=21​

7r3−6r4=1

Q. A uniform solid cone of height $H$ is cut at a distance $\times$ below its vertex, parallel to its base and the upper part is discarded. The $C . M .$ of the remaining part is found to be $50%$ lower, w.r.t. to the base. The ratio, $r = \frac{x}{H}$ satisfies.

$7 r^{3} - 8 r^{4} = 1$

$r = \frac{1}{2 ^{1/3}}$

$r = \frac{1}{2}$

$7 r^{3} - 6 r^{4} = 1$