Q31. I = $\dfrac{\text{nE}}{\text{R} + \text{nr}}$; for n
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$\Rightarrow$ I = $\dfrac{\text{nE}}{\text{R} + \text{nr}}$
$\Rightarrow$ I(R $+$ nr) = nE
$\Rightarrow$ IR $+$ nIr = nE
$\Rightarrow$ IR = nE $-$ nIr
$\Rightarrow$ IR = n(E $-$ Ir)
$\therefore$ n = $\dfrac{\text{IR}}{\text{E} - \text{Ir}}$ Ans
$\Rightarrow$ I(R $+$ nr) = nE
$\Rightarrow$ IR $+$ nIr = nE
$\Rightarrow$ IR = nE $-$ nIr
$\Rightarrow$ IR = n(E $-$ Ir)
$\therefore$ n = $\dfrac{\text{IR}}{\text{E} - \text{Ir}}$ Ans
Q32. I = $\dfrac{\text{nE}}{\text{nR} + \text{r}}$; for r
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$\Rightarrow$ I = $\dfrac{\text{nE}}{\text{nR} + \text{r}}$
$\Rightarrow$ I(nR $+$ r) = nE
$\Rightarrow$ nIR $+$ Ir = nE
$\Rightarrow$ Ir = nE $-$ nIR
$\therefore$ r = $\dfrac{\text{nE} - \text{nIR}}{\text{I}}$ Ans
$\Rightarrow$ I(nR $+$ r) = nE
$\Rightarrow$ nIR $+$ Ir = nE
$\Rightarrow$ Ir = nE $-$ nIR
$\therefore$ r = $\dfrac{\text{nE} - \text{nIR}}{\text{I}}$ Ans
Q33. $\dfrac{1}{\text{f}} = \dfrac{1}{\text{v}} - \dfrac{1}{\text{u}}$; for u
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$\Rightarrow$ $\dfrac{1}{\text{f}} = \dfrac{1}{\text{v}} - \dfrac{1}{\text{u}}$
$\Rightarrow$ $\dfrac{1}{\text{f}} = \dfrac{\text{u} - \text{v}}{\text{vu}}$
$\Rightarrow$ vu = f(u $-$ v)
$\Rightarrow$ vu = uf $-$ vf
$\Rightarrow$ vf = uf $-$vu
$\Rightarrow$ vf = u(f $-$ v)
$\Rightarrow \dfrac{\text{vf}}{\text{f} - \text{v}}$ = u Ans
$\Rightarrow$ $\dfrac{1}{\text{f}} = \dfrac{\text{u} - \text{v}}{\text{vu}}$
$\Rightarrow$ vu = f(u $-$ v)
$\Rightarrow$ vu = uf $-$ vf
$\Rightarrow$ vf = uf $-$vu
$\Rightarrow$ vf = u(f $-$ v)
$\Rightarrow \dfrac{\text{vf}}{\text{f} - \text{v}}$ = u Ans
Q34. A = $2\pi\text{r}(\text{r} + \text{h})$; for h
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$\Rightarrow$ A = $2\pi\text{r}(\text{r} + \text{h})$
$\Rightarrow$ A = $2\pi\text{r}^{2} + 2\pi\text{rh}$
$\Rightarrow$ $\text{A} - 2\pi\text{r}^{2}$ = $2\pi\text{rh}$
$\Rightarrow$ $\dfrac{\text{A} - 2\pi\text{r}^{2}}{2\pi\text{r}}$ = h Ans
$\Rightarrow$ A = $2\pi\text{r}^{2} + 2\pi\text{rh}$
$\Rightarrow$ $\text{A} - 2\pi\text{r}^{2}$ = $2\pi\text{rh}$
$\Rightarrow$ $\dfrac{\text{A} - 2\pi\text{r}^{2}}{2\pi\text{r}}$ = h Ans
Q35. m = $4\sqrt{\dfrac{\text{a}}{\text{b} + \text{c}}}$; for b
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$\Rightarrow$ m = $4\sqrt{\dfrac{\text{a}}{\text{b} + \text{c}}}$
$\Rightarrow \text{m}^{2} = 16 \times \dfrac{\text{a}}{\text{b} + \text{c}}$
$\Rightarrow \text{m}^{2} = \dfrac{\text{16a}}{\text{b} + \text{c}}$
$\Rightarrow \text{m}^{2}(\text{b} + \text{c}) = 16a$
$\Rightarrow \text{m}^{2}\text{b} + \text{m}^{2}\text{c} = 16a$
$\Rightarrow \text{m}^{2}\text{b} = 16a - \text{m}^{2}\text{c}$
$\Rightarrow \text{b} = \dfrac{16a - \text{m}^{2}\text{c}}{\text{m}^{2}}$ Ans
$\Rightarrow \text{m}^{2} = 16 \times \dfrac{\text{a}}{\text{b} + \text{c}}$
$\Rightarrow \text{m}^{2} = \dfrac{\text{16a}}{\text{b} + \text{c}}$
$\Rightarrow \text{m}^{2}(\text{b} + \text{c}) = 16a$
$\Rightarrow \text{m}^{2}\text{b} + \text{m}^{2}\text{c} = 16a$
$\Rightarrow \text{m}^{2}\text{b} = 16a - \text{m}^{2}\text{c}$
$\Rightarrow \text{b} = \dfrac{16a - \text{m}^{2}\text{c}}{\text{m}^{2}}$ Ans
Q36. z = $\dfrac{\text{a} - \text{b}}{4\text{b}}$; for b
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$\Rightarrow$ z = $\dfrac{\text{a} - \text{b}}{4\text{b}}$
$\Rightarrow$ z(4b) = a - b
$\Rightarrow$ 4bz = a - b
$\Rightarrow$ 4bz $+$ b = a
$\Rightarrow$ b(4z $+$ 1) = a
$\therefore \text{b} = \dfrac{a}{4\text{z} + 1}$ Ans
$\Rightarrow$ z(4b) = a - b
$\Rightarrow$ 4bz = a - b
$\Rightarrow$ 4bz $+$ b = a
$\Rightarrow$ b(4z $+$ 1) = a
$\therefore \text{b} = \dfrac{a}{4\text{z} + 1}$ Ans
Q37. S = $\dfrac{\text{n}}{2}[2\text{a} + (\text{n - 1})\text{d}]$; for d
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$\Rightarrow$ S = $\dfrac{\text{n}}{2}[2\text{a} + (\text{n - 1})\text{d}]$
$\Rightarrow$ 2S = $\text{n}[2\text{a} + \text{nd - d}]$
$\Rightarrow$ 2S = $2\text{an} + \text{n}^{2}\text{d} - \text{nd}$
$\Rightarrow$ 2S $-$ 2an = $\text{n}^{2}\text{d} - \text{nd}$
$\Rightarrow$ 2S $-$ 2an = $\text{d}(\text{n}^{2} - \text{n})$
$\therefore \dfrac{2\text{S} - 2\text{an}}{\text{n}^{2} - \text{n}}$ = $\text{d}$ Ans
$\Rightarrow$ 2S = $\text{n}[2\text{a} + \text{nd - d}]$
$\Rightarrow$ 2S = $2\text{an} + \text{n}^{2}\text{d} - \text{nd}$
$\Rightarrow$ 2S $-$ 2an = $\text{n}^{2}\text{d} - \text{nd}$
$\Rightarrow$ 2S $-$ 2an = $\text{d}(\text{n}^{2} - \text{n})$
$\therefore \dfrac{2\text{S} - 2\text{an}}{\text{n}^{2} - \text{n}}$ = $\text{d}$ Ans
Q38. T = $\dfrac{1}{\text{r}}\sqrt{\dfrac{T}{\pi\text{d}}}$; for d
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$\Rightarrow$ T = $\dfrac{1}{\text{r}}\sqrt{\dfrac{T}{\pi\text{d}}}$
$\Rightarrow \text{T}^{2} = \dfrac{1}{\text{r}^{2}} \times {\dfrac{T}{\pi\text{d}}}$
$\Rightarrow \text{T}^{2} = \dfrac{\text{T}}{\text{r}^{2}\pi\text{d}}$
$\Rightarrow \text{r}^{2}\pi\text{d}\text{T}^{2} = \text{T}$
$\Rightarrow \text{d}(\text{r}^{2}\pi\text{T}^{2}) = \text{T}$
$\Rightarrow \text{d} = \dfrac{\text{T}}{\text{r}^{2}\pi\text{T}^{2}}$
$\therefore \text{d} = \dfrac{1}{\text{r}^{2}\pi\text{T}}$ Ans
$\Rightarrow \text{T}^{2} = \dfrac{1}{\text{r}^{2}} \times {\dfrac{T}{\pi\text{d}}}$
$\Rightarrow \text{T}^{2} = \dfrac{\text{T}}{\text{r}^{2}\pi\text{d}}$
$\Rightarrow \text{r}^{2}\pi\text{d}\text{T}^{2} = \text{T}$
$\Rightarrow \text{d}(\text{r}^{2}\pi\text{T}^{2}) = \text{T}$
$\Rightarrow \text{d} = \dfrac{\text{T}}{\text{r}^{2}\pi\text{T}^{2}}$
$\therefore \text{d} = \dfrac{1}{\text{r}^{2}\pi\text{T}}$ Ans
Q39. $\dfrac{\text{a} + 1}{\text{b}} + 2 = \dfrac{3\text{a}}{\text{b}}$; for a
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$\Rightarrow \dfrac{\text{a} + 1}{\text{b}} + 2 = \dfrac{3\text{a}}{\text{b}}$
$\Rightarrow \dfrac{\text{a} + 1 + 2\text{b}}{\text{b}} = \dfrac{3\text{a}}{\text{b}}$
$\Rightarrow \text{b}(\text{a} + 1 + 2\text{b}) = \text{b}(3\text{a})$
$\Rightarrow \text{ab} + \text{b} + 2\text{b}^{2} = 3\text{ab}$
$\Rightarrow \text{b} + 2\text{b}^{2} = 3\text{ab} - \text{ab}$
$\Rightarrow \text{b}(1 + 2\text{b}) = \text{a}(3\text{b} - \text{b})$
$\Rightarrow \dfrac{\text{b}(1 + 2\text{b})}{3\text{b} - \text{b}} = \text{a}$
$\Rightarrow \dfrac{\text{b}(1 + 2\text{b})}{\text{b}(3 - 1)} = \text{a}$
$\Rightarrow \dfrac{\text{b}(1 + 2\text{b})}{\text{b}(2)} = \text{a}$
$\therefore \dfrac{1 + 2\text{b}}{2} = \text{a}$ Ans
$\Rightarrow \dfrac{\text{a} + 1 + 2\text{b}}{\text{b}} = \dfrac{3\text{a}}{\text{b}}$
$\Rightarrow \text{b}(\text{a} + 1 + 2\text{b}) = \text{b}(3\text{a})$
$\Rightarrow \text{ab} + \text{b} + 2\text{b}^{2} = 3\text{ab}$
$\Rightarrow \text{b} + 2\text{b}^{2} = 3\text{ab} - \text{ab}$
$\Rightarrow \text{b}(1 + 2\text{b}) = \text{a}(3\text{b} - \text{b})$
$\Rightarrow \dfrac{\text{b}(1 + 2\text{b})}{3\text{b} - \text{b}} = \text{a}$
$\Rightarrow \dfrac{\text{b}(1 + 2\text{b})}{\text{b}(3 - 1)} = \text{a}$
$\Rightarrow \dfrac{\text{b}(1 + 2\text{b})}{\text{b}(2)} = \text{a}$
$\therefore \dfrac{1 + 2\text{b}}{2} = \text{a}$ Ans
Q40. m = $\dfrac{\text{xy} - \text{z}}{\text{x} - 1}$; for x
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$\Rightarrow$ m = $\dfrac{\text{xy} - \text{z}}{\text{x} - 1}$
$\Rightarrow$ m(x $-$ 1) = xy $-$ z
$\Rightarrow$ mx $-$ m = xy $-$ z
$\Rightarrow$ mx $-$ xy = m $-$ z
$\Rightarrow$ x(m $-$ y) = m $-$ z
$\therefore$ x = $\dfrac{\text{m} - \text{z}}{\text{m} - \text{y}}$ Ans
$\Rightarrow$ m(x $-$ 1) = xy $-$ z
$\Rightarrow$ mx $-$ m = xy $-$ z
$\Rightarrow$ mx $-$ xy = m $-$ z
$\Rightarrow$ x(m $-$ y) = m $-$ z
$\therefore$ x = $\dfrac{\text{m} - \text{z}}{\text{m} - \text{y}}$ Ans
Q41. $\dfrac{\text{x} - \text{y}}{\text{x} + \text{y}} = \text{z}$; for y
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$\Rightarrow \dfrac{\text{x} - \text{y}}{\text{x} + \text{y}} = \text{z}$
$\Rightarrow \text{x} - \text{y} = \text{z}(\text{x} + \text{y})$
$\Rightarrow \text{x} - \text{y} = \text{zx} + \text{zy}$
$\Rightarrow \text{x} - \text{zx} = \text{y} + \text{zy}$
$\Rightarrow \text{x}(1 - \text{z}) = \text{y}(1 + \text{z})$
$\therefore \dfrac{\text{x}(1 - \text{z})}{1 + \text{z}} = \text{y}$ Ans
$\Rightarrow \text{x} - \text{y} = \text{z}(\text{x} + \text{y})$
$\Rightarrow \text{x} - \text{y} = \text{zx} + \text{zy}$
$\Rightarrow \text{x} - \text{zx} = \text{y} + \text{zy}$
$\Rightarrow \text{x}(1 - \text{z}) = \text{y}(1 + \text{z})$
$\therefore \dfrac{\text{x}(1 - \text{z})}{1 + \text{z}} = \text{y}$ Ans
Q42. $\dfrac{\text{P} - 2\text{l}}{2} = \text{b}$; for l
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$\Rightarrow \dfrac{\text{P} - 2\text{l}}{2} = \text{b}$
$\Rightarrow \text{P} - 2\text{l} = 2\text{b}$
$\Rightarrow \text{P} - 2\text{b} = 2\text{l}$
$\therefore \dfrac{\text{P} - 2\text{b}}{2} = \text{l}$ Ans
$\Rightarrow \text{P} - 2\text{l} = 2\text{b}$
$\Rightarrow \text{P} - 2\text{b} = 2\text{l}$
$\therefore \dfrac{\text{P} - 2\text{b}}{2} = \text{l}$ Ans
Q43. a = $\sqrt{\dfrac{\text{x} + \text{b}}{\text{x} - \text{b}}}$; for x
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$\Rightarrow \text{a} = \sqrt{\dfrac{\text{x} + \text{b}}{\text{x} - \text{b}}}$
$\Rightarrow \text{a}^{2} = \dfrac{\text{x} + \text{b}}{\text{x} - \text{b}}$
$\Rightarrow \text{a}^{2}(\text{x} - \text{b}) = \text{x} + \text{b}$
$\Rightarrow \text{a}^{2}\text{x} - \text{x} = \text{a}^{2}\text{b} + \text{b}$
$\Rightarrow \text{x}(\text{a}^{2} - 1) = \text{b}(\text{a}^{2} + 1)$
$\therefore \text{x} = \dfrac{\text{b}(\text{a}^{2} + 1)}{\text{a}^{2} - 1}$ Ans
$\Rightarrow \text{a}^{2} = \dfrac{\text{x} + \text{b}}{\text{x} - \text{b}}$
$\Rightarrow \text{a}^{2}(\text{x} - \text{b}) = \text{x} + \text{b}$
$\Rightarrow \text{a}^{2}\text{x} - \text{x} = \text{a}^{2}\text{b} + \text{b}$
$\Rightarrow \text{x}(\text{a}^{2} - 1) = \text{b}(\text{a}^{2} + 1)$
$\therefore \text{x} = \dfrac{\text{b}(\text{a}^{2} + 1)}{\text{a}^{2} - 1}$ Ans
Q44. F = $\dfrac{\text{mv} - \text{mu}}{\text{t}}$; for u
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$\Rightarrow \text{F} = \dfrac{\text{mv} - \text{mu}}{\text{t}}$
$\Rightarrow \text{Ft} = \text{mv} - \text{mu}$
$\Rightarrow \text{Ft} - \text{mv} = \text{mu}$
$\therefore \dfrac{\text{Ft} - \text{mv}}{\text{m}} = \text{u}$ Ans
$\Rightarrow \text{Ft} = \text{mv} - \text{mu}$
$\Rightarrow \text{Ft} - \text{mv} = \text{mu}$
$\therefore \dfrac{\text{Ft} - \text{mv}}{\text{m}} = \text{u}$ Ans
Q45. V = $\pi(\text{R}^{2} - \text{r}^{2})\text{h}$; for r
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$\Rightarrow \text{V} = \pi(\text{R}^{2} - \text{r}^{2})\text{h}$
$\Rightarrow \text{V} = \pi\text{R}^{2} - \pi\text{r}^{2}\text{h}$
$\Rightarrow \pi\text{r}^{2}\text{h} = \pi\text{R}^{2} - \text{V}$
$\Rightarrow \text{r}^{2}(\pi\text{h}) = \pi\text{R}^{2} - \text{V}$
$\Rightarrow \text{r}^{2} = \dfrac{\pi\text{R}^{2} - \text{V}}{\pi\text{h}}$
$\therefore \text{r} = \sqrt{\dfrac{\pi\text{R}^{2} - \text{V}}{\pi\text{h}}}$ Ans
$\Rightarrow \text{V} = \pi\text{R}^{2} - \pi\text{r}^{2}\text{h}$
$\Rightarrow \pi\text{r}^{2}\text{h} = \pi\text{R}^{2} - \text{V}$
$\Rightarrow \text{r}^{2}(\pi\text{h}) = \pi\text{R}^{2} - \text{V}$
$\Rightarrow \text{r}^{2} = \dfrac{\pi\text{R}^{2} - \text{V}}{\pi\text{h}}$
$\therefore \text{r} = \sqrt{\dfrac{\pi\text{R}^{2} - \text{V}}{\pi\text{h}}}$ Ans