Question of The Day12-10-2020

The ratio of volume and curved surface area of a cone is $$2\sqrt5 : 3$$. If the difference between radius and height of the cone is 5 cm, then what will be the curved surface area (in cm2) of the cone?

Correct Answer : c ) $$50\sqrt5π$$

Explanation :

Let us assume r, h, and L be the radius, height and slant height of the cone.

According to the question,

$${Volume \ of \ cone \over Curved \ surface \ area \ of \ cone}={{1\over3 }πr^2h \over πrL}$$

$${{1\over3 }πr^2h \over πrL}={2\sqrt5\over 3}$$

$${rh\over3L}={2\sqrt5\over3}$$

$$rh=2\sqrt5L$$ …. (1)

Now, we know for a cone

r2+h2=L2…….(2)

According to the question,

r – h = 5 …… (3)

Squaring both sides, we get,

r2 + h2 – 2rh = 25 …. (4)

From equation 2 and 4

25 + 2rh = L2

Now, from equation 1

$$25+2×2\sqrt5L=L^2$$

$$L^2-4\sqrt5 L-25=0$$

$$L^2-5\sqrt5L+\sqrt5L-25=0$$

$$L(L-5\sqrt5)+\sqrt5(L-5\sqrt5)=0$$

$$L=5\sqrt5$$, and $$L=-\sqrt5\ (Not \ possible)$$

Thus, $$L=5\sqrt5$$

Putting value of L in equation 1

$$rh=2\sqrt5×5\sqrt5$$

⇒ rh = 50 …. (5)

From equation 3 and 5

Radius of cone = 10 cm and height of cone = 5 cm

Therefore,

Curved surface area of cone=$$πrL$$

$$π×10×5\sqrt5=50\sqrt5π$$

Hence, (c) is the correct answer.

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS, UPSC NDA etc.

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