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Question Number 16514 Answers: 1 Comments: 1

Help plz Given 2d^2 x/dt^2 +2dx/dt+x=ksint where k is constant show that if x=n and dx/dt=0 when t=0 then x=e^(−1/2t) {(n+2/5k)cos(t/2)+)+(n+4/5k)sin(t/2)} −1/5ksint−2/5kcost.

$${Help}\:{plz} \\ $$$${Given}\:\mathrm{2}{d}^{\mathrm{2}} {x}/{dt}^{\mathrm{2}} +\mathrm{2}{dx}/{dt}+{x}={ksint} \\ $$$${where}\:{k}\:{is}\:{constant}\:{show}\:{that}\:{if}\:{x}={n} \\ $$$${and}\:{dx}/{dt}=\mathrm{0}\:{when}\:{t}=\mathrm{0}\:{then}\: \\ $$$$\left.{x}={e}^{−\mathrm{1}/\mathrm{2}{t}} \left\{\left({n}+\mathrm{2}/\mathrm{5}{k}\right){cos}\left({t}/\mathrm{2}\right)+\right)+\left({n}+\mathrm{4}/\mathrm{5}{k}\right){sin}\left({t}/\mathrm{2}\right)\right\} \\ $$$$−\mathrm{1}/\mathrm{5}{ksint}−\mathrm{2}/\mathrm{5}{kcost}. \\ $$

Question Number 16506 Answers: 0 Comments: 0

A train 1 moves from east to west (clockwise) and another train 2 moves from west to east (anticlockwise) on the equator with equal speeds relative to ground. The ratio of their centripetal acceleration (a_1 /a_2 ) relative to centre of earth is (1) > 1 (2) < 1

$$\mathrm{A}\:\mathrm{train}\:\mathrm{1}\:\mathrm{moves}\:\mathrm{from}\:\mathrm{east}\:\mathrm{to}\:\mathrm{west} \\ $$$$\left(\mathrm{clockwise}\right)\:\mathrm{and}\:\mathrm{another}\:\mathrm{train}\:\mathrm{2}\:\mathrm{moves} \\ $$$$\mathrm{from}\:\mathrm{west}\:\mathrm{to}\:\mathrm{east}\:\left(\mathrm{anticlockwise}\right)\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{equator}\:\mathrm{with}\:\mathrm{equal}\:\mathrm{speeds}\:\mathrm{relative} \\ $$$$\mathrm{to}\:\mathrm{ground}.\:\mathrm{The}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{their} \\ $$$$\mathrm{centripetal}\:\mathrm{acceleration}\:\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{2}} }\:\mathrm{relative}\:\mathrm{to} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{earth}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:>\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:<\:\mathrm{1} \\ $$

Question Number 16505 Answers: 1 Comments: 0

What is the maximum magnitude of change in velocity of a motorcycle moving with a uniform speed v_0 in a circular path of length l = (π/3)R and radius R? Treat motorcycle as a particle (1) ∣Δv^→ ∣ = (v_0 /2) (2) ∣Δv^→ ∣ = v_0

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{magnitude}\:\mathrm{of} \\ $$$$\mathrm{change}\:\mathrm{in}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{motorcycle} \\ $$$$\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{speed}\:{v}_{\mathrm{0}} \:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{circular}\:\mathrm{path}\:\mathrm{of}\:\mathrm{length}\:{l}\:=\:\frac{\pi}{\mathrm{3}}{R}\:\mathrm{and} \\ $$$$\mathrm{radius}\:\mathrm{R}?\:\mathrm{Treat}\:\mathrm{motorcycle}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{particle} \\ $$$$\left(\mathrm{1}\right)\:\mid\Delta\overset{\rightarrow} {{v}}\mid\:=\:\frac{{v}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\mid\Delta\overset{\rightarrow} {{v}}\mid\:=\:{v}_{\mathrm{0}} \\ $$

Question Number 16504 Answers: 2 Comments: 0

An object moves in a circular path with a constant speed in the xy plane with the centre at the origin. When the object is at x = −2 m, its velocity is −(4 m/s)j^∧ . Then objects velocity at y = 2 m is (1) 4 m/s i^∧ (2) (−4 m/s) i^∧ Using this data, find objects acceleration when it is at y = 2 m (1) 8 m/s^2 i^∧ (2) −8 m/s^2 j^∧

$$\mathrm{An}\:\mathrm{object}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{xy}\:\mathrm{plane}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{at}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{When}\:\mathrm{the} \\ $$$$\mathrm{object}\:\mathrm{is}\:\mathrm{at}\:{x}\:=\:−\mathrm{2}\:\mathrm{m},\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{is} \\ $$$$−\left(\mathrm{4}\:\mathrm{m}/\mathrm{s}\right)\overset{\wedge} {{j}}.\:\mathrm{Then}\:\mathrm{objects}\:\mathrm{velocity}\:\mathrm{at} \\ $$$${y}\:=\:\mathrm{2}\:\mathrm{m}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{4}\:\mathrm{m}/\mathrm{s}\:\overset{\wedge} {{i}} \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{4}\:\mathrm{m}/\mathrm{s}\right)\:\overset{\wedge} {{i}} \\ $$$$\mathrm{Using}\:\mathrm{this}\:\mathrm{data},\:\mathrm{find}\:\mathrm{objects}\:\mathrm{acceleration} \\ $$$$\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at}\:{y}\:=\:\mathrm{2}\:\mathrm{m} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\overset{\wedge} {{i}} \\ $$$$\left(\mathrm{2}\right)\:−\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\overset{\wedge} {{j}} \\ $$

Question Number 16502 Answers: 1 Comments: 0

The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector. (True/False)

$$\mathrm{The}\:\mathrm{acceleration}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{in} \\ $$$$\mathrm{uniform}\:\mathrm{circular}\:\mathrm{motion}\:\mathrm{averaged}\:\mathrm{over} \\ $$$$\mathrm{one}\:\mathrm{cycle}\:\mathrm{is}\:\mathrm{a}\:\mathrm{null}\:\mathrm{vector}.\:\left(\mathrm{True}/\mathrm{False}\right) \\ $$

Question Number 16501 Answers: 1 Comments: 0

3+3

$$\mathrm{3}+\mathrm{3} \\ $$

Question Number 16499 Answers: 1 Comments: 0

A particle is moving in parabolic path x^2 = y, with constant speed u. Find the acceleration of the particle when it crossess origin. Also find the radius of curvature at origin.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{parabolic}\:\mathrm{path} \\ $$$${x}^{\mathrm{2}} \:=\:{y},\:\mathrm{with}\:\mathrm{constant}\:\mathrm{speed}\:{u}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{when}\:\mathrm{it} \\ $$$$\mathrm{crossess}\:\mathrm{origin}.\:\mathrm{Also}\:\mathrm{find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of} \\ $$$$\mathrm{curvature}\:\mathrm{at}\:\mathrm{origin}. \\ $$

Question Number 16493 Answers: 0 Comments: 1

Find L{tcosh2t}

$${Find}\:{L}\left\{{tcosh}\mathrm{2}{t}\right\} \\ $$

Question Number 16492 Answers: 1 Comments: 0

solve Mdc/dt=P+km−Cm where M.P.m and k are constant solve the equation given C=a when t=o

$${solve} \\ $$$${Mdc}/{dt}={P}+{km}−{Cm}\:{where}\:{M}.{P}.{m}\:{and} \\ $$$${k}\:{are}\:{constant} \\ $$$${solve}\:{the}\:{equation}\:{given}\:{C}={a}\:{when}\:{t}={o} \\ $$

Question Number 16491 Answers: 0 Comments: 0

Men are running in a line along a road with velocity 9 km/hr behind one another at equal distances of 20 m. Cyclists are also riding along the same line in the same direction at 18 km/hr at equal intervals of 30 m. The speed with which an observer must travel along the road in opposite direction of so that whenever he meets a runner he also meets a cyclist is (1) 9 km/h (2) 12 km/h (3) 18 km/h (4) 6 km/h

$$\mathrm{Men}\:\mathrm{are}\:\mathrm{running}\:\mathrm{in}\:\mathrm{a}\:\mathrm{line}\:\mathrm{along}\:\mathrm{a}\:\mathrm{road} \\ $$$$\mathrm{with}\:\mathrm{velocity}\:\mathrm{9}\:\mathrm{km}/\mathrm{hr}\:\mathrm{behind}\:\mathrm{one} \\ $$$$\mathrm{another}\:\mathrm{at}\:\mathrm{equal}\:\mathrm{distances}\:\mathrm{of}\:\mathrm{20}\:\mathrm{m}. \\ $$$$\mathrm{Cyclists}\:\mathrm{are}\:\mathrm{also}\:\mathrm{riding}\:\mathrm{along}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{line}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{direction}\:\mathrm{at}\:\mathrm{18}\:\mathrm{km}/\mathrm{hr} \\ $$$$\mathrm{at}\:\mathrm{equal}\:\mathrm{intervals}\:\mathrm{of}\:\mathrm{30}\:\mathrm{m}.\:\mathrm{The}\:\mathrm{speed} \\ $$$$\mathrm{with}\:\mathrm{which}\:\mathrm{an}\:\mathrm{observer}\:\mathrm{must}\:\mathrm{travel} \\ $$$$\mathrm{along}\:\mathrm{the}\:\mathrm{road}\:\mathrm{in}\:\mathrm{opposite}\:\mathrm{direction}\:\mathrm{of} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{whenever}\:\mathrm{he}\:\mathrm{meets}\:\mathrm{a}\:\mathrm{runner}\:\mathrm{he} \\ $$$$\mathrm{also}\:\mathrm{meets}\:\mathrm{a}\:\mathrm{cyclist}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{9}\:\mathrm{km}/\mathrm{h} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{12}\:\mathrm{km}/\mathrm{h} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{18}\:\mathrm{km}/\mathrm{h} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{6}\:\mathrm{km}/\mathrm{h} \\ $$

Question Number 16497 Answers: 1 Comments: 0

Two particles are revolving on two coplanar circles with constant angular velocities ω_1 and ω_2 respectively. Their time periods are T_1 and T_2 then prove that the time taken by second particle to complete one revolution more than the first particle, T, is given by T = ((T_1 T_2 )/(T_1 − T_2 ))

$$\mathrm{Two}\:\mathrm{particles}\:\mathrm{are}\:\mathrm{revolving}\:\mathrm{on}\:\mathrm{two} \\ $$$$\mathrm{coplanar}\:\mathrm{circles}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{angular} \\ $$$$\mathrm{velocities}\:\omega_{\mathrm{1}} \:\mathrm{and}\:\omega_{\mathrm{2}} \:\mathrm{respectively}.\:\mathrm{Their} \\ $$$$\mathrm{time}\:\mathrm{periods}\:\mathrm{are}\:{T}_{\mathrm{1}} \:\mathrm{and}\:{T}_{\mathrm{2}} \:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{second}\:\mathrm{particle} \\ $$$$\mathrm{to}\:\mathrm{complete}\:\mathrm{one}\:\mathrm{revolution}\:\mathrm{more}\:\mathrm{than} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{particle},\:{T},\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${T}\:=\:\frac{{T}_{\mathrm{1}} {T}_{\mathrm{2}} }{{T}_{\mathrm{1}} \:−\:{T}_{\mathrm{2}} } \\ $$

Question Number 16489 Answers: 1 Comments: 0

Rain is falling vertically with a speed of 4 m/s. After some time, wind starts blowing with a speed of 3 m/s in the north to south direction. In order to protect himself from rain, a man standing on the ground should hold his umbrella at an angle θ given by (1) θ = tan^(−1) 3/4 with the vertical towards south (2) θ = tan^(−1) 3/4 with the vertical towards north (3) θ = cot^(−1) 3/4 with the vertical towards south (1) θ = cot^(−1) 3/4 with the vertical towards north

$$\mathrm{Rain}\:\mathrm{is}\:\mathrm{falling}\:\mathrm{vertically}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed} \\ $$$$\mathrm{of}\:\mathrm{4}\:\mathrm{m}/\mathrm{s}.\:\mathrm{After}\:\mathrm{some}\:\mathrm{time},\:\mathrm{wind}\:\mathrm{starts} \\ $$$$\mathrm{blowing}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{3}\:\mathrm{m}/\mathrm{s}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{north}\:\mathrm{to}\:\mathrm{south}\:\mathrm{direction}.\:\mathrm{In}\:\mathrm{order}\:\mathrm{to} \\ $$$$\mathrm{protect}\:\mathrm{himself}\:\mathrm{from}\:\mathrm{rain},\:\mathrm{a}\:\mathrm{man} \\ $$$$\mathrm{standing}\:\mathrm{on}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{should}\:\mathrm{hold}\:\mathrm{his} \\ $$$$\mathrm{umbrella}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{given}\:\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} \mathrm{3}/\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{vertical} \\ $$$$\mathrm{towards}\:\mathrm{south} \\ $$$$\left(\mathrm{2}\right)\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} \mathrm{3}/\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{vertical} \\ $$$$\mathrm{towards}\:\mathrm{north} \\ $$$$\left(\mathrm{3}\right)\:\theta\:=\:\mathrm{cot}^{−\mathrm{1}} \mathrm{3}/\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{vertical} \\ $$$$\mathrm{towards}\:\mathrm{south} \\ $$$$\left(\mathrm{1}\right)\:\theta\:=\:\mathrm{cot}^{−\mathrm{1}} \mathrm{3}/\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{vertical} \\ $$$$\mathrm{towards}\:\mathrm{north} \\ $$

Question Number 16485 Answers: 1 Comments: 0

If tan^2 θ=2 tan^2 φ+1, then cos 2θ+sin^2 φ equals

$$\mathrm{If}\:\mathrm{tan}^{\mathrm{2}} \theta=\mathrm{2}\:\mathrm{tan}^{\mathrm{2}} \phi+\mathrm{1},\:\mathrm{then}\:\mathrm{cos}\:\mathrm{2}\theta+\mathrm{sin}^{\mathrm{2}} \phi \\ $$$$\mathrm{equals} \\ $$

Question Number 16483 Answers: 0 Comments: 12

Question Number 16475 Answers: 1 Comments: 0

use L{f′(t)} to find laplace transform of d^2 x/dt^2 +n^2 x=kcoswt given that x=0 and dx/dt=0 when t=0 PLZ HELP.

$${use}\:{L}\left\{{f}'\left({t}\right)\right\}\:{to}\:{find}\:{laplace}\: \\ $$$${transform}\:{of}\:{d}^{\mathrm{2}} {x}/{dt}^{\mathrm{2}} +{n}^{\mathrm{2}} {x}={kcoswt} \\ $$$${given}\:{that}\:{x}=\mathrm{0}\:{and}\:{dx}/{dt}=\mathrm{0}\:{when}\:{t}=\mathrm{0} \\ $$$${PLZ}\:{HELP}. \\ $$$$ \\ $$

Question Number 16466 Answers: 1 Comments: 0

Question Number 16464 Answers: 0 Comments: 2

Question Number 17493 Answers: 0 Comments: 0

What is logarithm series?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{logarithm}\:\mathrm{series}? \\ $$

Question Number 16457 Answers: 1 Comments: 0

Find the range of f(x) = (3/(2 − x^2 ))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}\:−\:{x}^{\mathrm{2}} } \\ $$

Question Number 16455 Answers: 2 Comments: 0

The speed of a projectile when it is at its greatest height is (√(2/5)) times its speed at half the maximum height. What is its angle of projection?

$$\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at} \\ $$$$\mathrm{its}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{is}\:\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\:\mathrm{times}\:\mathrm{its} \\ $$$$\mathrm{speed}\:\mathrm{at}\:\mathrm{half}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{its}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{projection}? \\ $$

Question Number 16451 Answers: 0 Comments: 0

The sixth term of an AP is 2, and its common difference is greater than one. The value of the common difference of the progression so that the product of the first, fourth and fifth terms is greatest is

$$\mathrm{The}\:\mathrm{sixth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{is}\:\mathrm{2},\:\mathrm{and}\:\mathrm{its} \\ $$$$\mathrm{common}\:\mathrm{difference}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{one}. \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{progression}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{first},\:\mathrm{fourth}\:\mathrm{and}\:\mathrm{fifth}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{greatest}\:\mathrm{is} \\ $$

Question Number 16430 Answers: 1 Comments: 2

In ΔABC with usual notation (r_1 /(bc)) + (r_2 /(ca)) + (r_3 /(ab)) is (1) (1/r) − (1/R) (2) (1/r) − (1/(2R)) (3) (1/r) + (1/(2R)) (4) (1/r) + (1/R)

$$\mathrm{In}\:\Delta{ABC}\:\mathrm{with}\:\mathrm{usual}\:\mathrm{notation} \\ $$$$\frac{{r}_{\mathrm{1}} }{{bc}}\:+\:\frac{{r}_{\mathrm{2}} }{{ca}}\:+\:\frac{{r}_{\mathrm{3}} }{{ab}}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{r}}\:−\:\frac{\mathrm{1}}{{R}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}}{{r}}\:−\:\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{1}}{{r}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}}{{r}}\:+\:\frac{\mathrm{1}}{{R}} \\ $$

Question Number 17647 Answers: 0 Comments: 5

ABC is a triangular park with AB = AC = 100 m. A clock tower is situated at the midpoint of BC. The angles of elevation of top of the tower at A and B are cot^(−1) (3.2) and cosec^(−1) (2.6) respectively. The height of tower is

$${ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangular}\:\mathrm{park}\:\mathrm{with}\:{AB}\:= \\ $$$${AC}\:=\:\mathrm{100}\:\mathrm{m}.\:\mathrm{A}\:\mathrm{clock}\:\mathrm{tower}\:\mathrm{is}\:\mathrm{situated} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{BC}.\:\mathrm{The}\:\mathrm{angles}\:\mathrm{of} \\ $$$$\mathrm{elevation}\:\mathrm{of}\:\mathrm{top}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tower}\:\mathrm{at}\:{A}\:\mathrm{and} \\ $$$${B}\:\mathrm{are}\:\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{3}.\mathrm{2}\right)\:\mathrm{and}\:\mathrm{cosec}^{−\mathrm{1}} \left(\mathrm{2}.\mathrm{6}\right) \\ $$$$\mathrm{respectively}.\:\mathrm{The}\:\mathrm{height}\:\mathrm{of}\:\mathrm{tower}\:\mathrm{is} \\ $$

Question Number 16419 Answers: 1 Comments: 0

3 cubes of metal whose edges are 3,4 and 5 respectively are melted and formed into a single cube. If there be no loss of metal in the process find the side of the new cube.

$$\mathrm{3}\:\mathrm{cubes}\:\mathrm{of}\:\mathrm{metal}\:\mathrm{whose}\:\mathrm{edges}\:\mathrm{are}\:\mathrm{3},\mathrm{4} \\ $$$$\mathrm{and}\:\mathrm{5}\:\mathrm{respectively}\:\mathrm{are}\:\mathrm{melted}\:\mathrm{and} \\ $$$$\mathrm{formed}\:\mathrm{into}\:\mathrm{a}\:\mathrm{single}\:\mathrm{cube}.\:\mathrm{If}\:\mathrm{there}\:\mathrm{be} \\ $$$$\mathrm{no}\:\mathrm{loss}\:\mathrm{of}\:\mathrm{metal}\:\mathrm{in}\:\mathrm{the}\:\mathrm{process}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{new}\:\mathrm{cube}. \\ $$

Question Number 16418 Answers: 1 Comments: 0

A solid sphere of radius 4 cm is melted and recast into ′n′ solid hemispheres of radius 2 cm each. Find n.

$$\mathrm{A}\:\mathrm{solid}\:\mathrm{sphere}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{4}\:\mathrm{cm}\:\mathrm{is}\:\mathrm{melted} \\ $$$$\mathrm{and}\:\mathrm{recast}\:\mathrm{into}\:'{n}'\:\mathrm{solid}\:\mathrm{hemispheres}\:\mathrm{of} \\ $$$$\mathrm{radius}\:\mathrm{2}\:\mathrm{cm}\:\mathrm{each}.\:\mathrm{Find}\:{n}. \\ $$

Question Number 16409 Answers: 3 Comments: 9

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