A man inside an elevator uses weighing machine to weigh himself. With what acceleration should the elevator descend so that the weighing machine reports the weight of the man to be half of its true value?
The reading of the weighing machine depends on the force by which the machine is pressed. Let us say the machine is pressed downwards by force N. The machine will exert an equal upward force N on the man.
Let m be the mass of the man. For the man system,
mg – N = ma
$mg – \frac{{mg}}{2} = ma$
$ \Rightarrow a = \frac{g}{2}$
Consider a particle in circular motion as shown in the figure. The position vector sweeps equal area in equal time. Select correct option(s).
(A) the speed is constantContinue reading Circular Motion Radius Area →
A uniform rope of linear mass density $\lambda $ is used to release block m with uniform acceleration a. Find the tension at a point P on the rope at a distance l from the block as shown in the figure.
Solution
Let mass m and the rope of length l above the block be the system.
Mass of rope of length $l = \lambda l$
For the system under consideration, $(m + \lambda l)g – {T_P} = (m + \lambda l)a$
$ \Rightarrow {T_P} = (m + \lambda l)(g – a)$
The given expression can be written as,
$\frac{1}{2}\left[ {(1 – \cos {2^\circ }) + (1 – \cos {4^\circ }) + ……………. + (1 – \cos {{360}^\circ })} \right]$
$ = \frac{1}{2}\left[ {(1 + 1 + …….. + 1) – (\cos {2^\circ } + \cos {4^\circ } + ……….. + \cos {{360}^\circ })} \right]$
There are 180 terms in the series.
$ = \frac{1}{2}\left[ {180 – \frac{{\cos \left( {{2^\circ } + \frac{{180 – 1}}{2} \times {2^\circ }} \right)\sin \left( {\frac{{180 \times {2^\circ }}}{2}} \right)}}{{\sin \left( {\frac{{{2^\circ }}}{2}} \right)}}} \right]$
$ = \frac{1}{2}\left( {180 – \frac{{\cos {{181}^\circ }\sin {{180}^\circ }}}{{\sin {1^\circ }}}} \right) = 90$
LHS = ${\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{\{ 1 + (\sin \theta – i\cos \theta )\} (\sin \theta + i\cos \theta )}}} \right]^n}$
$ = {\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{(\sin \theta + i\cos \theta ) + \{ {{\sin }^2}\theta – {{(i\cos \theta )}^2}\} }}} \right]^n}$
$ = {\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{(\sin \theta + i\cos \theta ) + ({{\sin }^2}\theta + {{\cos }^2}\theta )}}} \right]^n}$
$ = {\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{(1 + \sin \theta + i\cos \theta )}}} \right]^n}$
$ = {(\sin \theta + i\cos \theta )^n}$
$ = {\left\{ {\cos \left( {\frac{\pi }{2} – \theta } \right) + i\sin \left( {\frac{\pi }{2} – \theta } \right)} \right\}^n}$
$ = \cos n\left( {\frac{\pi }{2} – \theta } \right) + i\sin n\left( {\frac{\pi }{2} – \theta } \right)$
= RHS
The given expression can be written as,
$({a^2}{\sec ^2}\theta – {a^2}) + {a^2} + ({b^2}{\rm{cose}}{{\rm{c}}^2}\theta – {b^2}) + {b^2}$
$ = {a^2}{\tan ^2}\theta + {b^2}{\cot ^2}\theta + {a^2} + {b^2}$
Using A.M. $\geq $ G.M.,
$\frac{{{a^2}{{\tan }^2}\theta + {b^2}{{\cot }^2}\theta }}{2} \ge \sqrt {{a^2}{{\tan }^2}\theta .{b^2}{{\cot }^2}\theta } $
$ \Rightarrow {a^2}{\tan ^2}\theta + {b^2}{\cot ^2}\theta \ge 2|ab|$
$ \Rightarrow {a^2}{\tan ^2}\theta + {b^2}{\cot ^2}\theta + {a^2} + {b^2} \ge 2|ab| + {a^2} + {b^2}$
$\therefore{({a^2}{\sec ^2}\theta + {b^2}{\rm{cose}}{{\rm{c}}^2}\theta )_{\min }} = {a^2} + {b^2} + 2|ab|$
The given limit = ${e^{\mathop {\lim }\limits_{x \to 0} \frac{{\tan x – \sin x}}{{{{\sin }^3}x}}}}$
$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\cos x}} – 1}}{{{{\sin }^2}x}}}}$
$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{\cos x.{{\sin }^2}x}}}}$
$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{\cos x.(1 – {{\cos }^2}x)}}}}$
$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x.(1 + \cos x)}}}}$
$ = {e^{1/2}} = \sqrt e $
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