Q46. s = $\text{u} + \dfrac{1}{2}a(2\text{t} - 1)$; for a
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$\Rightarrow \text{s} = \text{u} + \dfrac{1}{2}a(2\text{t} - 1)$
$\Rightarrow \text{s} - \text{u} = \dfrac{1}{2}a(2\text{t} - 1)$
$\Rightarrow 2(\text{s} - \text{u}) = a(2\text{t} - 1)$
$\therefore \dfrac{2(\text{s} - \text{u})}{2\text{t} - 1} = a$ Ans
$\Rightarrow \text{s} - \text{u} = \dfrac{1}{2}a(2\text{t} - 1)$
$\Rightarrow 2(\text{s} - \text{u}) = a(2\text{t} - 1)$
$\therefore \dfrac{2(\text{s} - \text{u})}{2\text{t} - 1} = a$ Ans
Q47. s = $\text{u} + \dfrac{1}{2}a(2\text{t} - 1)$; for t
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$\Rightarrow \text{s} = \text{u} + \dfrac{1}{2}a(2\text{t} - 1)$
$\Rightarrow \text{s} - \text{u} = \dfrac{1}{2}a(2\text{t} - 1)$
$\Rightarrow 2(\text{s} - \text{u}) = a(2\text{t} - 1)$
$\Rightarrow 2\text{s} - 2\text{u} = 2\text{at} - a$
$\Rightarrow 2\text{s} - 2\text{u} + a = 2\text{at}$
$\therefore \dfrac{2\text{s} - 2\text{u} + a}{2\text{a}} = \text{t}$ Ans
$\Rightarrow \text{s} - \text{u} = \dfrac{1}{2}a(2\text{t} - 1)$
$\Rightarrow 2(\text{s} - \text{u}) = a(2\text{t} - 1)$
$\Rightarrow 2\text{s} - 2\text{u} = 2\text{at} - a$
$\Rightarrow 2\text{s} - 2\text{u} + a = 2\text{at}$
$\therefore \dfrac{2\text{s} - 2\text{u} + a}{2\text{a}} = \text{t}$ Ans
Q48. a = $\sqrt{\dfrac{15\text{x} + 16\text{y}}{\text{x} + \text{y}}}$; for y
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$\Rightarrow \text{a} = \sqrt{\dfrac{15\text{x} + 16\text{y}}{\text{x} + \text{y}}}$
$\Rightarrow \text{a}^{2} = \dfrac{15\text{x} + 16\text{y}}{\text{x} + \text{y}}$
$\Rightarrow \text{a}^{2}(\text{x} + \text{y}) = 15\text{x} + 16\text{y}$
$\Rightarrow \text{a}^{2}\text{x} + \text{a}^{2}\text{y} = 15\text{x} + 16\text{y}$
$\Rightarrow \text{a}^{2}\text{x} - 15\text{x} = 16\text{y} - \text{a}^{2}\text{y}$
$\Rightarrow \text{x}(\text{a}^{2} - 15) = \text{y}(16 - \text{a}^{2})$
$\therefore \dfrac{\text{x}(\text{a}^{2} - 15)}{16 - \text{a}^{2}} = \text{y}$ Ans
$\Rightarrow \text{a}^{2} = \dfrac{15\text{x} + 16\text{y}}{\text{x} + \text{y}}$
$\Rightarrow \text{a}^{2}(\text{x} + \text{y}) = 15\text{x} + 16\text{y}$
$\Rightarrow \text{a}^{2}\text{x} + \text{a}^{2}\text{y} = 15\text{x} + 16\text{y}$
$\Rightarrow \text{a}^{2}\text{x} - 15\text{x} = 16\text{y} - \text{a}^{2}\text{y}$
$\Rightarrow \text{x}(\text{a}^{2} - 15) = \text{y}(16 - \text{a}^{2})$
$\therefore \dfrac{\text{x}(\text{a}^{2} - 15)}{16 - \text{a}^{2}} = \text{y}$ Ans
Q49. $\text{C} = 2\pi(\text{R} - \text{r})$; for r
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$\Rightarrow \text{C} = 2\pi(\text{R} - \text{r})$
$\Rightarrow \text{C} = 2\pi\text{R} - 2\pi\text{r}$
$\Rightarrow 2\pi\text{r} = 2\pi\text{R} - \text{C}$
$\therefore \text{r} = \dfrac{2\pi\text{R} - \text{C}}{2\pi}$ Ans
$\Rightarrow \text{C} = 2\pi\text{R} - 2\pi\text{r}$
$\Rightarrow 2\pi\text{r} = 2\pi\text{R} - \text{C}$
$\therefore \text{r} = \dfrac{2\pi\text{R} - \text{C}}{2\pi}$ Ans
Q50. $\text{s} = \text{ut} + \dfrac{1}{2}\text{at}^{2}$; for a
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$\Rightarrow \text{s} = \text{ut} + \dfrac{1}{2}\text{at}^{2}$
$\Rightarrow \text{s} - \text{ut} = \dfrac{1}{2}\text{at}^{2}$
$\Rightarrow 2(\text{s} - \text{ut}) = \text{at}^{2}$
$\therefore \dfrac{2(\text{s} - \text{ut})}{\text{at}^{2}} = \text{a}$ Ans
$\Rightarrow \text{s} - \text{ut} = \dfrac{1}{2}\text{at}^{2}$
$\Rightarrow 2(\text{s} - \text{ut}) = \text{at}^{2}$
$\therefore \dfrac{2(\text{s} - \text{ut})}{\text{at}^{2}} = \text{a}$ Ans
Q51. $\text{t} = 2\pi\sqrt{\dfrac{l}{\text{g}}}$; for g
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$\Rightarrow \text{t} = 2\pi\sqrt{\dfrac{l}{\text{g}}}$
$\Rightarrow \text{t}^{2} = 4\pi^{2} \times \dfrac{l}{\text{g}}$
$\Rightarrow \text{t}^{2} = \dfrac{4\pi^{2}l}{\text{g}}$
$\Rightarrow \text{t}^{2}\text{g} = 4\pi^{2}l$
$\therefore \text{g} = \dfrac{4\pi^{2}l}{\text{t}^{2}}$ Ans
$\Rightarrow \text{t}^{2} = 4\pi^{2} \times \dfrac{l}{\text{g}}$
$\Rightarrow \text{t}^{2} = \dfrac{4\pi^{2}l}{\text{g}}$
$\Rightarrow \text{t}^{2}\text{g} = 4\pi^{2}l$
$\therefore \text{g} = \dfrac{4\pi^{2}l}{\text{t}^{2}}$ Ans
Q52. $l = \text{a} + (\text{n} - 1)\text{d}$; for n
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$\Rightarrow l = \text{a} + (\text{n} - 1)\text{d}$
$\Rightarrow l = \text{a} + \text{nd} - \text{d}$
$\Rightarrow l - \text{a} = \text{nd} - \text{d}$
$\Rightarrow l - \text{a} + \text{d} = \text{nd}$
$\therefore \dfrac{l - \text{a} + \text{d}}{\text{d}} = \text{n}$ Ans
$\Rightarrow l = \text{a} + \text{nd} - \text{d}$
$\Rightarrow l - \text{a} = \text{nd} - \text{d}$
$\Rightarrow l - \text{a} + \text{d} = \text{nd}$
$\therefore \dfrac{l - \text{a} + \text{d}}{\text{d}} = \text{n}$ Ans
Q53. $\text{S} = \dfrac{n}{2}[2\text{a} + (\text{n} - 1)\text{d}]$; for d
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$\Rightarrow \text{S} = \dfrac{n}{2}[2\text{a} + (\text{n} - 1)\text{d}]$
$\Rightarrow \text{S} = \dfrac{n}{2}[2\text{a} + \text{nd} - \text{d}]$
$\Rightarrow 2\text{S} = \text{n}[2\text{a} + \text{nd} - \text{d}]$
$\Rightarrow 2\text{S} = 2\text{an} + \text{n}^{2}\text{d} - \text{nd}$
$\Rightarrow 2\text{S} - 2\text{an} = \text{n}^{2}\text{d} - \text{nd}$
$\Rightarrow 2(\text{S} - \text{an}) = \text{d}(\text{n}^{2} - \text{n})$
$\Rightarrow \dfrac{2(\text{S} - \text{an})}{\text{n}^{2} - \text{n}} = \text{d}$
$\therefore \dfrac{2(\text{S} - \text{an})}{\text{n}(\text{n} - 1)} = \text{d}$ Ans
$\Rightarrow \text{S} = \dfrac{n}{2}[2\text{a} + \text{nd} - \text{d}]$
$\Rightarrow 2\text{S} = \text{n}[2\text{a} + \text{nd} - \text{d}]$
$\Rightarrow 2\text{S} = 2\text{an} + \text{n}^{2}\text{d} - \text{nd}$
$\Rightarrow 2\text{S} - 2\text{an} = \text{n}^{2}\text{d} - \text{nd}$
$\Rightarrow 2(\text{S} - \text{an}) = \text{d}(\text{n}^{2} - \text{n})$
$\Rightarrow \dfrac{2(\text{S} - \text{an})}{\text{n}^{2} - \text{n}} = \text{d}$
$\therefore \dfrac{2(\text{S} - \text{an})}{\text{n}(\text{n} - 1)} = \text{d}$ Ans
Q54. $\text{I} = \dfrac{nE}{\text{nR} + \text{r}}$; for n
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$\Rightarrow \text{I} = \dfrac{nE}{\text{nR} + \text{r}}$
$\Rightarrow \text{InR} + \text{Ir} = \text{nE}$
$\Rightarrow \text{Ir} = \text{nE} - \text{InR}$
$\Rightarrow \text{Ir} = \text{n}(\text{E} - \text{IR})$
$\therefore \dfrac{\text{Ir}}{\text{E} - \text{IR}} = \text{n}$ Ans
$\Rightarrow \text{InR} + \text{Ir} = \text{nE}$
$\Rightarrow \text{Ir} = \text{nE} - \text{InR}$
$\Rightarrow \text{Ir} = \text{n}(\text{E} - \text{IR})$
$\therefore \dfrac{\text{Ir}}{\text{E} - \text{IR}} = \text{n}$ Ans
Q55. $\text{F} = \dfrac{\text{mv} - \text{mu}}{\text{t}}$; for u
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$\Rightarrow \text{F} = \dfrac{\text{mv} - \text{mu}}{\text{t}}$
$\Rightarrow \text{Ft} = \text{mv} - \text{mu}$
$\Rightarrow \text{mu} = \text{mv} - \text{Ft}$
$\therefore \text{u} = \dfrac{\text{mv} - \text{Ft}}{\text{m}}$ Ans
$\Rightarrow \text{Ft} = \text{mv} - \text{mu}$
$\Rightarrow \text{mu} = \text{mv} - \text{Ft}$
$\therefore \text{u} = \dfrac{\text{mv} - \text{Ft}}{\text{m}}$ Ans
Q56. $\text{F} = \dfrac{9}{5}\text{C} + 32$; for C
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$\Rightarrow \text{F} = \dfrac{9}{5}\text{C} + 32$
$\Rightarrow \dfrac{5}{9}\text{F} = \text{C} + 32$
$\therefore \dfrac{5}{9}\text{F} - 32 = \text{C}$ Ans
$\Rightarrow \dfrac{5}{9}\text{F} = \text{C} + 32$
$\therefore \dfrac{5}{9}\text{F} - 32 = \text{C}$ Ans
Q57. $\dfrac{\text{a} + 1}{\text{b}} + 3 = \dfrac{4a}{\text{b}}$; for b
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$\Rightarrow \dfrac{\text{a} + 1}{\text{b}} + 3 = \dfrac{4\text{a}}{\text{b}}$
$\Rightarrow 3 = \dfrac{4\text{a}}{\text{b}} - \dfrac{\text{a} + 1}{\text{b}}$
$\Rightarrow 3 = \dfrac{4\text{a} - \text{a} + 1}{\text{b}}$
$\Rightarrow 3\text{b} = 4\text{a} - \text{a} + 1$
$\Rightarrow 3\text{b} = 3\text{a} + 1$
$\therefore \text{b} = \dfrac{3\text{a} + 1}{3}$ Ans
$\Rightarrow 3 = \dfrac{4\text{a}}{\text{b}} - \dfrac{\text{a} + 1}{\text{b}}$
$\Rightarrow 3 = \dfrac{4\text{a} - \text{a} + 1}{\text{b}}$
$\Rightarrow 3\text{b} = 4\text{a} - \text{a} + 1$
$\Rightarrow 3\text{b} = 3\text{a} + 1$
$\therefore \text{b} = \dfrac{3\text{a} + 1}{3}$ Ans
Q58. $\text{V} = \pi(\text{R}^{2} - r^{2})\text{h}$; for r
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$\Rightarrow \text{V} = \pi(\text{R}^{2} - r^{2})\text{h}$
$\Rightarrow \text{V} = (\text{R}^{2}\pi - r^{2}\pi)\text{h}$
$\Rightarrow \text{V} = \text{R}^{2}\pi\text{h} - r^{2}\pi\text{h}$
$\Rightarrow r^{2}\pi\text{h} = \text{R}^{2}\pi\text{h} - \text{V}$
$\Rightarrow r^{2} = \dfrac{\text{R}^{2}\pi\text{h}}{\pi\text{h}} - \dfrac{\text{V}}{\pi\text{h}}$
$\Rightarrow r^{2} = \text{R}^{2} - \dfrac{\text{V}}{\pi\text{h}}$
$\therefore r = \sqrt{\text{R}^{2} - \dfrac{\text{V}}{\pi\text{h}}}$ Ans
$\Rightarrow \text{V} = (\text{R}^{2}\pi - r^{2}\pi)\text{h}$
$\Rightarrow \text{V} = \text{R}^{2}\pi\text{h} - r^{2}\pi\text{h}$
$\Rightarrow r^{2}\pi\text{h} = \text{R}^{2}\pi\text{h} - \text{V}$
$\Rightarrow r^{2} = \dfrac{\text{R}^{2}\pi\text{h}}{\pi\text{h}} - \dfrac{\text{V}}{\pi\text{h}}$
$\Rightarrow r^{2} = \text{R}^{2} - \dfrac{\text{V}}{\pi\text{h}}$
$\therefore r = \sqrt{\text{R}^{2} - \dfrac{\text{V}}{\pi\text{h}}}$ Ans
Q59. $\text{x} = \dfrac{3\text{a} - 5\text{b}}{3\text{a} + 5\text{b}}$; for b
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$\Rightarrow \text{x} = \dfrac{3\text{a} - 5\text{b}}{3\text{a} + 5\text{b}}$
$\Rightarrow \text{x}(3\text{a} + 5\text{b}) = 3\text{a} - 5\text{b}$
$\Rightarrow 5\text{bx} + 5\text{b} = 3\text{a} - 3\text{ax}$
$\Rightarrow 5\text{b}(\text{x} + 1) = 3\text{a}(1 - \text{x})$
$\therefore \text{b} = \dfrac{3\text{a}(1 - \text{x})}{5(\text{x} + 1)}$ Ans
$\Rightarrow \text{x}(3\text{a} + 5\text{b}) = 3\text{a} - 5\text{b}$
$\Rightarrow 5\text{bx} + 5\text{b} = 3\text{a} - 3\text{ax}$
$\Rightarrow 5\text{b}(\text{x} + 1) = 3\text{a}(1 - \text{x})$
$\therefore \text{b} = \dfrac{3\text{a}(1 - \text{x})}{5(\text{x} + 1)}$ Ans
Q60. $\text{a} = -\dfrac{\text{bc}}{\text{b} + \text{c}}$; for c
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$\Rightarrow \text{a} = -\dfrac{\text{bc}}{\text{b} + \text{c}}$
$\Rightarrow \text{a}(\text{b} + \text{c}) = -\text{bc}$
$\Rightarrow \text{ab} + \text{ac} = -\text{bc}$
$\Rightarrow \text{ac} + \text{bc} = -\text{ab}$
$\Rightarrow \text{c}(\text{a} + \text{b}) = -\text{ab}$
$\therefore \text{c} = \dfrac{-\text{ab}}{\text{a} + \text{b}}$ Ans
$\Rightarrow \text{a}(\text{b} + \text{c}) = -\text{bc}$
$\Rightarrow \text{ab} + \text{ac} = -\text{bc}$
$\Rightarrow \text{ac} + \text{bc} = -\text{ab}$
$\Rightarrow \text{c}(\text{a} + \text{b}) = -\text{ab}$
$\therefore \text{c} = \dfrac{-\text{ab}}{\text{a} + \text{b}}$ Ans