The volume of a cone is $V = \frac{1}{3} π r^{2} h$ . Make r as the subject of the formula.

$r = \sqrt{\frac{3 V^{2}}{π h}}$

$r = \sqrt{\frac{V}{3 π h}}$

$r = \sqrt{\frac{3 V}{π h}}$

$r = \sqrt{\frac{V}{3 π h}}$

Video Solution

Text Solution

Verified by Experts

The correct Answer is:C

(i) apply transposition
(ii) Cross multiply and make $r^{2}$ as the subject.
(iii) Apply square root on both sides.

Updated on:21/07/2023

Class 8 MATHS QUESTION BANK

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Similar Practice Problems

Question 1 - Select One
The volume of a cone is $V = \frac{1}{3} π r^{2} h$ . Make r as the subject of the formula.
A $r = \sqrt{\frac{3 V^{2}}{π h}}$
B $r = \sqrt{\frac{V}{3 π h}}$
C $r = \sqrt{\frac{3 V}{π h}}$
D $r = \sqrt{\frac{V}{3 π h}}$
Question 1 - Select One
The volume of a sphere is given by $V = \frac{4}{3} π r^{3}$ Make r as the subject of the formula.
A $\sqrt{\frac{3 V}{4 π}}$
B $\frac{3 V}{4 π}$
C ${(\frac{3 V}{π})}^{\frac{1}{3}}$
D ${(\frac{3 V}{4 π})}^{\frac{1}{3}}$
Question 1 - Select One
If $S = \frac{a}{1 - r}$ , then make r as the subject of the formula.
A $r = \frac{a}{S} + 1$
B $r = 1 + \frac{S}{a}$
C $r = 1 - \frac{S}{a}$
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If $S = \frac{a}{1 - r}$ , then make r as the subject of the formula.
A $r = \frac{a}{S} + 1$
B $r = 1 + \frac{S}{a}$
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Question 1 - Select One
The volume of a sphere is given by $V = \frac{4}{3} π r^{3}$ Make r as the subject of the formula.
A $\sqrt{\frac{3 V}{4 π}}$
B $\frac{3 V}{4 π}$
C ${(\frac{3 V}{π})}^{\frac{1}{3}}$
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Question 1 - Select One
In $S = π \sqrt{r^{2} + h^{2}}$ , make h as the subject of the formula.
A $h = \sqrt{\frac{S}{π r} - r^{2}}$
B $h = \sqrt{\frac{S^{2}}{π r^{2}} + r^{2}}$
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Question 1 - Select One
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A $b = \frac{2 + H}{H a}$
B $b = \frac{H a}{2 a - H}$
C $b = \frac{2 a - H}{H a}$
D $b = \frac{H a}{2 a + H}$

The volume of a cone is $V = \frac{1}{3} π r^{2} h$ . Make r as the subject of the formula.

The correct Answer is:C

(i) apply transposition
(ii) Cross multiply and make $r^{2}$ as the subject.
(iii) Apply square root on both sides.

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The volume of a cone is $V = \frac{1}{3} π r^{2} h$ . Make r as the subject of the formula.

The volume of a sphere is given by $V = \frac{4}{3} π r^{3}$ Make r as the subject of the formula.

If $S = \frac{a}{1 - r}$ , then make r as the subject of the formula.

If $S = \frac{a}{1 - r}$ , then make r as the subject of the formula.

The volume of a sphere is given by $V = \frac{4}{3} π r^{3}$ Make r as the subject of the formula.

In $S = π \sqrt{r^{2} + h^{2}}$ , make h as the subject of the formula.

If $H = \frac{2 b}{a + b}$ , then make b as the subject of the formula.

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PEARSON IIT JEE FOUNDATION-Formulae-Level 2

The volume of a cone is V=13πr2h. Make r as the subject of the formula.

The correct Answer is:C

(i) apply transposition (ii) Cross multiply and make r2 as the subject. (iii) Apply square root on both sides.

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The volume of a cone is V=13πr2h. Make r as the subject of the formula.

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The volume of a cone is $V = \frac{1}{3} π r^{2} h$ . Make r as the subject of the formula.

(i) apply transposition
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(iii) Apply square root on both sides.

The volume of a cone is $V = \frac{1}{3} π r^{2} h$ . Make r as the subject of the formula.

The volume of a sphere is given by $V = \frac{4}{3} π r^{3}$ Make r as the subject of the formula.

If $S = \frac{a}{1 - r}$ , then make r as the subject of the formula.

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