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Question Number 24459 Answers: 2 Comments: 0

If the sides of a rectangle are increased each by 10%, find the % increase in its diagonal.

$$\mathrm{If}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{are}\:\mathrm{increased} \\ $$$$\mathrm{each}\:\mathrm{by}\:\mathrm{10\%},\:\mathrm{find}\:\mathrm{the}\:\%\:\mathrm{increase}\:\mathrm{in}\:\mathrm{its} \\ $$$$\mathrm{diagonal}. \\ $$

Question Number 24456 Answers: 1 Comments: 0

The product of a monomial by a binomial is a

$${The}\:{product}\:{of}\:{a}\:{monomial}\:{by}\:{a}\:{binomial}\:{is}\:{a} \\ $$

Question Number 24434 Answers: 0 Comments: 0

Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities ω (anti-clockwise) and ω (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by (3R + δ). They are now brought in contact (δ → 0). (a) Show the frictional forces just after contact. (b) Identify forces and torques external to the system just after contact. (c) What would be the ratio of final angular velocities when friction ceases?

$$\mathrm{Two}\:\mathrm{cylindrical}\:\mathrm{hollow}\:\mathrm{drums}\:\mathrm{of}\:\mathrm{radii} \\ $$$${R}\:\mathrm{and}\:\mathrm{2}{R},\:\mathrm{and}\:\mathrm{of}\:\mathrm{a}\:\mathrm{common}\:\mathrm{height}\:{h}, \\ $$$$\mathrm{are}\:\mathrm{rotating}\:\mathrm{with}\:\mathrm{angular}\:\mathrm{velocities}\:\omega \\ $$$$\left(\mathrm{anti}-\mathrm{clockwise}\right)\:\mathrm{and}\:\omega\:\left(\mathrm{clockwise}\right), \\ $$$$\mathrm{respectively}.\:\mathrm{Their}\:\mathrm{axes},\:\mathrm{fixed}\:\mathrm{are} \\ $$$$\mathrm{parallel}\:\mathrm{and}\:\mathrm{in}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{plane} \\ $$$$\mathrm{separated}\:\mathrm{by}\:\left(\mathrm{3}{R}\:+\:\delta\right).\:\mathrm{They}\:\mathrm{are}\:\mathrm{now} \\ $$$$\mathrm{brought}\:\mathrm{in}\:\mathrm{contact}\:\left(\delta\:\rightarrow\:\mathrm{0}\right). \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{the}\:\mathrm{frictional}\:\mathrm{forces}\:\mathrm{just}\:\mathrm{after} \\ $$$$\mathrm{contact}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Identify}\:\mathrm{forces}\:\mathrm{and}\:\mathrm{torques}\:\mathrm{external} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{system}\:\mathrm{just}\:\mathrm{after}\:\mathrm{contact}. \\ $$$$\left(\mathrm{c}\right)\:\mathrm{What}\:\mathrm{would}\:\mathrm{be}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{final} \\ $$$$\mathrm{angular}\:\mathrm{velocities}\:\mathrm{when}\:\mathrm{friction}\:\mathrm{ceases}? \\ $$

Question Number 24436 Answers: 0 Comments: 0

For a reversible reaction A ⇌ B. Find K_(eq) at 2727°C temperature. Given : Δ_r H° = −30 kJ mol^(−1) (at 2727°C) Δ_r S° = 10 JK^(−1) (at 2727°C) R = 8.314 JK^(−1) mol^(−1)

$$\mathrm{For}\:\mathrm{a}\:\mathrm{reversible}\:\mathrm{reaction}\:\mathrm{A}\:\rightleftharpoons\:\mathrm{B}.\:\mathrm{Find} \\ $$$$\mathrm{K}_{\mathrm{eq}} \:\mathrm{at}\:\mathrm{2727}°\mathrm{C}\:\mathrm{temperature}. \\ $$$$\mathrm{Given}\::\:\Delta_{\mathrm{r}} \mathrm{H}°\:=\:−\mathrm{30}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} \:\left(\mathrm{at}\:\mathrm{2727}°\mathrm{C}\right) \\ $$$$\Delta_{\mathrm{r}} \mathrm{S}°\:=\:\mathrm{10}\:\mathrm{JK}^{−\mathrm{1}} \:\left(\mathrm{at}\:\mathrm{2727}°\mathrm{C}\right) \\ $$$$\mathrm{R}\:=\:\mathrm{8}.\mathrm{314}\:\mathrm{JK}^{−\mathrm{1}} \:\mathrm{mol}^{−\mathrm{1}} \\ $$

Question Number 24428 Answers: 1 Comments: 0

In a race on a track of a certain length, A beats B by 20 m. When A was 10 m ahead of the mid−point of the track B was 2 m behind it. Find the length of the track (in m).

$$\mathrm{In}\:\mathrm{a}\:\mathrm{race}\:\mathrm{on}\:\mathrm{a}\:\mathrm{track}\:\:\mathrm{of}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{length}, \\ $$$$\mathrm{A}\:\:\mathrm{beats}\:\mathrm{B}\:\mathrm{by}\:\mathrm{20}\:\mathrm{m}.\:\mathrm{When}\:\mathrm{A}\:\mathrm{was}\:\mathrm{10}\:\mathrm{m} \\ $$$$\mathrm{ahead}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mid}−\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{track}\:\mathrm{B} \\ $$$$\mathrm{was}\:\mathrm{2}\:\mathrm{m}\:\mathrm{behind}\:\mathrm{it}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{track}\:\left(\mathrm{in}\:\mathrm{m}\right). \\ $$

Question Number 24426 Answers: 1 Comments: 0

A number is multiplied by 2(1/3) times itself and then 61 is subtracted from the product obtained. If the final result is 9200, then the number is

$$\mathrm{A}\:\mathrm{number}\:\mathrm{is}\:\mathrm{multiplied}\:\mathrm{by}\:\mathrm{2}\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{times} \\ $$$$\mathrm{itself}\:\mathrm{and}\:\mathrm{then}\:\mathrm{61}\:\mathrm{is}\:\mathrm{subtracted}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{product}\:\mathrm{obtained}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{final}\:\mathrm{result} \\ $$$$\mathrm{is}\:\mathrm{9200},\:\mathrm{then}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is} \\ $$

Question Number 24447 Answers: 0 Comments: 8

Figure shows two discs of same mass m. They are rigidly attached to a spring of stiffness k. The system is in equilibrium. From this equilibrium position, the upper disc is pressed down slowly by a distance x and released. Find the minimum value of x, if the lower disc is just lifted off the ground.

$${Figure}\:{shows}\:{two}\:{discs}\:{of}\:{same}\:{mass} \\ $$$${m}.\:{They}\:{are}\:{rigidly}\:{attached}\:{to}\:{a}\:{spring} \\ $$$${of}\:{stiffness}\:{k}.\:{The}\:{system}\:{is}\:{in} \\ $$$${equilibrium}.\:{From}\:{this}\:{equilibrium} \\ $$$${position},\:{the}\:{upper}\:{disc}\:{is}\:{pressed} \\ $$$${down}\:{slowly}\:{by}\:{a}\:{distance}\:{x}\:{and} \\ $$$${released}.\:{Find}\:{the}\:{minimum}\:{value}\:{of} \\ $$$${x},\:{if}\:{the}\:{lower}\:{disc}\:{is}\:{just}\:{lifted}\:{off} \\ $$$${the}\:{ground}. \\ $$

Question Number 24443 Answers: 2 Comments: 0

Solve for x: (10^(−4) x)^x =4×10^(−8)

$${Solve}\:{for}\:{x}: \\ $$$$\left(\mathrm{10}^{−\mathrm{4}} {x}\right)^{{x}} =\mathrm{4}×\mathrm{10}^{−\mathrm{8}} \\ $$

Question Number 24417 Answers: 1 Comments: 0

The time taken by a train l metres long running at x km/h to pass a man who is running at y km/h in the direction opposite to that of the train = The time taken to cover l metres at _____ km/h.

$$\mathrm{The}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{a}\:\mathrm{train}\:{l}\:\mathrm{metres}\:\mathrm{long} \\ $$$$\mathrm{running}\:\mathrm{at}\:{x}\:\mathrm{km}/\mathrm{h}\:\mathrm{to}\:\mathrm{pass}\:\mathrm{a}\:\mathrm{man}\:\mathrm{who} \\ $$$$\mathrm{is}\:\mathrm{running}\:\mathrm{at}\:{y}\:\mathrm{km}/\mathrm{h}\:\mathrm{in}\:\mathrm{the}\:\mathrm{direction} \\ $$$$\mathrm{opposite}\:\mathrm{to}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the}\:\mathrm{train}\:=\:\mathrm{The}\:\mathrm{time} \\ $$$$\mathrm{taken}\:\mathrm{to}\:\mathrm{cover}\:{l}\:\mathrm{metres}\:\mathrm{at}\:\_\_\_\_\_\:\mathrm{km}/\mathrm{h}. \\ $$

Question Number 24438 Answers: 0 Comments: 2

Find the sum to infinite terms of the series (x/(1−x^2 ))+(x^2 /(1−x^4 ))+(x^4 /(1−x^8 ))+......

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinite}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{series}\:\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} }+\frac{{x}^{\mathrm{4}} }{\mathrm{1}−{x}^{\mathrm{8}} }+...... \\ $$

Question Number 24411 Answers: 1 Comments: 0

Question Number 24405 Answers: 0 Comments: 0

Find the horizontal assymptot lim_(x→∞ ) (((3x^2 +4))^(1/6) /((1−2x^3 ))^(1/9) )

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{assymptot} \\ $$$$\underset{{x}\rightarrow\infty\:} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{6}}]{\mathrm{3x}^{\mathrm{2}} +\mathrm{4}}}{\sqrt[{\mathrm{9}}]{\mathrm{1}−\mathrm{2x}^{\mathrm{3}} }} \\ $$$$ \\ $$

Question Number 24404 Answers: 0 Comments: 0

put a_n = (1/(2n))∙(3/(2n))∙(5/(2n))∙ ... ∙((2n−1)/(2n))∙e^n lim_(n→∞) a_n = ?

$${put}\:{a}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}{n}}\centerdot\frac{\mathrm{3}}{\mathrm{2}{n}}\centerdot\frac{\mathrm{5}}{\mathrm{2}{n}}\centerdot\:...\:\centerdot\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}\centerdot{e}^{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {{lim}a}_{{n}} =\:? \\ $$$$ \\ $$

Question Number 24387 Answers: 1 Comments: 0

Two particle move along an x−axis. The position of particle 1 is given; by x=6.00t^2 +3.00t+2.00((m/s)) the acceleration of particle 2 is given by a=−8.00t((m/s^2 )) and,at t=0,its velocity is 20 ((m/s)).when the velocities of the particles match, what is their velocity? plzz help

Question Number 24379 Answers: 2 Comments: 0

Question Number 24371 Answers: 0 Comments: 5

Two blocks are moving together under the action of a constant horizontal external force F. If the smaller block is at rest with respect to the bigger block due to the friction between them, then the normal reaction between the bigger block and floor is

$$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{are}\:\mathrm{moving}\:\mathrm{together}\:\mathrm{under} \\ $$$$\mathrm{the}\:\mathrm{action}\:\mathrm{of}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{horizontal} \\ $$$$\mathrm{external}\:\mathrm{force}\:{F}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{block}\:\mathrm{is} \\ $$$$\mathrm{at}\:\mathrm{rest}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bigger}\:\mathrm{block} \\ $$$$\mathrm{due}\:\mathrm{to}\:\mathrm{the}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{them},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{normal}\:\mathrm{reaction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{bigger} \\ $$$$\mathrm{block}\:\mathrm{and}\:\mathrm{floor}\:\mathrm{is} \\ $$

Question Number 24369 Answers: 0 Comments: 2

Question Number 24366 Answers: 1 Comments: 1

Given the 7-element set A = {a, b, c, d, e, f, g}, find a collection T of 3- element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{7}-\mathrm{element}\:\mathrm{set}\:{A}\:=\:\left\{{a},\:{b},\:{c},\right. \\ $$$$\left.{d},\:{e},\:{f},\:{g}\right\},\:\mathrm{find}\:\mathrm{a}\:\mathrm{collection}\:{T}\:\mathrm{of}\:\mathrm{3}- \\ $$$$\mathrm{element}\:\mathrm{subsets}\:\mathrm{of}\:{A}\:\mathrm{such}\:\mathrm{that}\:\mathrm{each} \\ $$$$\mathrm{pair}\:\mathrm{of}\:\mathrm{elements}\:\mathrm{from}\:{A}\:\mathrm{occurs}\:\mathrm{exactly} \\ $$$$\mathrm{in}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{subsets}\:\mathrm{of}\:{T}. \\ $$

Question Number 24363 Answers: 3 Comments: 0

∫(ae)^x dx

$$\int\left({ae}\right)^{{x}} {dx} \\ $$

Question Number 24360 Answers: 0 Comments: 1

Question Number 24359 Answers: 1 Comments: 1

Question Number 24362 Answers: 0 Comments: 1

if the points P,Q(x,7),R,S(6,y) in this order divide the line segment joining A(2,p) and B(7,10) in 5 equal parts find x,y, and p.

$${if}\:{the}\:{points}\:{P},{Q}\left({x},\mathrm{7}\right),{R},{S}\left(\mathrm{6},{y}\right)\:{in}\:{this}\:{order}\:{divide}\:{the}\:{line}\:{segment}\:{joining}\:{A}\left(\mathrm{2},{p}\right)\:{and}\:{B}\left(\mathrm{7},\mathrm{10}\right)\:{in}\:\mathrm{5}\:{equal}\:{parts}\:{find}\:{x},{y},\:{and}\:{p}. \\ $$

Question Number 24355 Answers: 1 Comments: 0

(√(1+(√(4+(√(16+(√(256.....))))))))=?

$$\sqrt{\mathrm{1}+\sqrt{\mathrm{4}+\sqrt{\mathrm{16}+\sqrt{\mathrm{256}.....}}}}=? \\ $$

Question Number 24332 Answers: 0 Comments: 0

z^(−4_(=1/3(1−(√(3i)))) ) ?

$$\mathrm{z}^{−\mathrm{4}_{=\mathrm{1}/\mathrm{3}\left(\mathrm{1}−\sqrt{\left.\mathrm{3i}\right)}\right.} } ? \\ $$

Question Number 24303 Answers: 1 Comments: 0

Assertion: Enthalpy of combustion is negative. Reason: Combustion reaction can be exothermic or endothermic.

$$\boldsymbol{\mathrm{Assertion}}:\:\mathrm{Enthalpy}\:\mathrm{of}\:\mathrm{combustion}\:\mathrm{is} \\ $$$$\mathrm{negative}. \\ $$$$\boldsymbol{\mathrm{Reason}}:\:\mathrm{Combustion}\:\mathrm{reaction}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{exothermic}\:\mathrm{or}\:\mathrm{endothermic}. \\ $$

Question Number 24301 Answers: 0 Comments: 4

In the figure shown below, all surfaces are smooth, strings and pulley are ideal. If the wedge is moving with acceleration a towards the right, then the acceleration of the block with respect to the wedge at that instant is

$$\mathrm{In}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{shown}\:\mathrm{below},\:\mathrm{all}\:\mathrm{surfaces} \\ $$$$\mathrm{are}\:\mathrm{smooth},\:\mathrm{strings}\:\mathrm{and}\:\mathrm{pulley}\:\mathrm{are}\:\mathrm{ideal}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{wedge}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{acceleration} \\ $$$${a}\:\mathrm{towards}\:\mathrm{the}\:\mathrm{right},\:\mathrm{then}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{block}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{wedge} \\ $$$$\mathrm{at}\:\mathrm{that}\:\mathrm{instant}\:\mathrm{is} \\ $$

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