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Question Number 18440 Answers: 0 Comments: 0

Question Number 18428 Answers: 0 Comments: 0

Question Number 18415 Answers: 1 Comments: 1

Calculate the magnetic field produced at ground level by a 15A current flowing in a long horizontal wire suspended at a height of 7.5m

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{magnetic}\:\mathrm{field}\:\mathrm{produced}\:\mathrm{at}\:\mathrm{ground}\:\mathrm{level}\:\mathrm{by}\:\mathrm{a}\:\mathrm{15A}\:\mathrm{current} \\ $$$$\mathrm{flowing}\:\mathrm{in}\:\mathrm{a}\:\mathrm{long}\:\mathrm{horizontal}\:\mathrm{wire}\:\mathrm{suspended}\:\mathrm{at}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\mathrm{7}.\mathrm{5m} \\ $$

Question Number 18411 Answers: 1 Comments: 0

A glass bulb contains 2.24 L of H_2 and 1.12 L of D_2 at S.T.P. It is connected to a fully evacuated bulb by a stopcock with a small opening. The stopcock is opened for sometime and then closed. The first bulb now contains 0.1 g of D_2 . Calculate the percentage composition by weight of the gases in the second bulb.

$$\mathrm{A}\:\mathrm{glass}\:\mathrm{bulb}\:\mathrm{contains}\:\mathrm{2}.\mathrm{24}\:\mathrm{L}\:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} \:\mathrm{and} \\ $$$$\mathrm{1}.\mathrm{12}\:\mathrm{L}\:\mathrm{of}\:\mathrm{D}_{\mathrm{2}} \:\mathrm{at}\:\mathrm{S}.\mathrm{T}.\mathrm{P}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{to} \\ $$$$\mathrm{a}\:\mathrm{fully}\:\mathrm{evacuated}\:\mathrm{bulb}\:\mathrm{by}\:\mathrm{a}\:\mathrm{stopcock} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{small}\:\mathrm{opening}.\:\mathrm{The}\:\mathrm{stopcock}\:\mathrm{is} \\ $$$$\mathrm{opened}\:\mathrm{for}\:\mathrm{sometime}\:\mathrm{and}\:\mathrm{then}\:\mathrm{closed}. \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{bulb}\:\mathrm{now}\:\mathrm{contains}\:\mathrm{0}.\mathrm{1}\:\mathrm{g}\:\mathrm{of}\:\mathrm{D}_{\mathrm{2}} . \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{percentage}\:\mathrm{composition} \\ $$$$\mathrm{by}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{the}\:\mathrm{gases}\:\mathrm{in}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{bulb}. \\ $$

Question Number 20955 Answers: 1 Comments: 0

lemme join miss Tawa Tawa here. It takes 8 painters working at the same rate ,5 hours to paint a house.If 6 painters are working at 2/3 the rate of the 8 painters,how long would it take them to paint the same house?

$$\mathrm{lemme}\:\mathrm{join}\:\mathrm{miss}\:\mathrm{Tawa}\:\mathrm{Tawa}\:\mathrm{here}. \\ $$$$ \\ $$$$\mathrm{It}\:\mathrm{takes}\:\mathrm{8}\:\mathrm{painters}\:\mathrm{working}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{rate}\:,\mathrm{5}\:\mathrm{hours}\:\mathrm{to}\:\mathrm{paint}\:\mathrm{a} \\ $$$$\mathrm{house}.\mathrm{If}\:\mathrm{6}\:\mathrm{painters}\:\mathrm{are}\:\mathrm{working}\:\mathrm{at} \\ $$$$\mathrm{2}/\mathrm{3}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{the}\:\mathrm{8}\:\mathrm{painters},\mathrm{how} \\ $$$$\mathrm{long}\:\mathrm{would}\:\mathrm{it}\:\mathrm{take}\:\mathrm{them}\:\mathrm{to}\:\mathrm{paint} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{house}? \\ $$

Question Number 18429 Answers: 0 Comments: 6

30cm^3 of hydrogen at s.t.p combines with 20cm^3 of oxygen to form steam according to the following equation, 2H_2 (g) + O_2 (g) → 2H_2 O (g). Calculate the total volume of gaseous mixture at the end of the reaction.

$$\mathrm{30cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{hydrogen}\:\mathrm{at}\:\mathrm{s}.\mathrm{t}.\mathrm{p}\:\mathrm{combines}\:\mathrm{with}\:\mathrm{20cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{oxygen}\:\mathrm{to}\:\mathrm{form}\:\mathrm{steam}\: \\ $$$$\mathrm{according}\:\mathrm{to}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation},\:\:\mathrm{2H}_{\mathrm{2}} \:\left(\mathrm{g}\right)\:+\:\mathrm{O}_{\mathrm{2}} \:\left(\mathrm{g}\right)\:\rightarrow\:\mathrm{2H}_{\mathrm{2}} \mathrm{O}\:\left(\mathrm{g}\right). \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{total}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{gaseous}\:\mathrm{mixture}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{reaction}. \\ $$

Question Number 18384 Answers: 1 Comments: 0

Let a, b, c ∈ R, a ≠ 0, such that a and 4a + 3b + 2c have the same sign. Show that the equation ax^2 + bx + c = 0 can not have both roots in the interval (1, 2).

$$\mathrm{Let}\:{a},\:{b},\:{c}\:\in\:{R},\:{a}\:\neq\:\mathrm{0},\:\mathrm{such}\:\mathrm{that}\:{a}\:\mathrm{and} \\ $$$$\mathrm{4}{a}\:+\:\mathrm{3}{b}\:+\:\mathrm{2}{c}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{sign}.\:\mathrm{Show} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{can} \\ $$$$\mathrm{not}\:\mathrm{have}\:\mathrm{both}\:\mathrm{roots}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\left(\mathrm{1},\:\mathrm{2}\right). \\ $$

Question Number 19200 Answers: 0 Comments: 5

A river of width d is flowing with speed u as shown in the figure. John can swim with maximum speed v relative to the river and can cross it in shortest time T. John starts at A. B is the point directly opposite to A on the other bank of the river. If t be the time John takes to reach the opposite bank, match the situation in the column I to the possibilities in column II. Column I (A) John reaches to the left of B (B) John reaches to the right of B (C) John reaches the point B (D) John drifts along the bank while minimizing the time Column II (p) t = T (q) t > T (r) u < v (s) u > v

Question Number 18357 Answers: 0 Comments: 0

From the topic transformer prove that: e = (√2) ε cos(ωt)

$$\mathrm{From}\:\mathrm{the}\:\mathrm{topic}\:\mathrm{transformer} \\ $$$$ \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\mathrm{e}\:=\:\sqrt{\mathrm{2}}\:\varepsilon\:\mathrm{cos}\left(\omega\mathrm{t}\right) \\ $$

Question Number 18349 Answers: 0 Comments: 0

Consider the iteration x_(k+1) =x_k −(([f(x)]^2 )/(f(x_k +f(x_k ))−f(x_k ))), k=0,1,2,... for the solution of f(x)=0. Explain the connection with Newton′s method, and show that (x_k ) converges quadratically if x_0 is sufficiently close to the solution.

$${Consider}\:{the}\:{iteration} \\ $$$${x}_{{k}+\mathrm{1}} ={x}_{{k}} −\frac{\left[{f}\left({x}\right)\right]^{\mathrm{2}} }{{f}\left({x}_{{k}} +{f}\left({x}_{{k}} \right)\right)−{f}\left({x}_{{k}} \right)},\:\:\:\:\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},... \\ $$$${for}\:{the}\:{solution}\:{of}\:{f}\left({x}\right)=\mathrm{0}.\:{Explain}\:{the} \\ $$$${connection}\:{with}\:{Newton}'{s}\:{method},\:{and}\:{show} \\ $$$${that}\:\left({x}_{{k}} \right)\:{converges}\:{quadratically}\:{if}\:{x}_{\mathrm{0}} \:{is} \\ $$$${sufficiently}\:{close}\:{to}\:{the}\:{solution}. \\ $$$$ \\ $$

Question Number 18366 Answers: 1 Comments: 0

Question Number 18322 Answers: 1 Comments: 1

The pulley arrangements are identical. The mass of the rope is negligible. In (a), the mass m is lifted up by attaching a mass (2m) to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2mg. In which case, the acceleration of m is more?

$$\mathrm{The}\:\mathrm{pulley}\:\mathrm{arrangements}\:\mathrm{are}\:\mathrm{identical}. \\ $$$$\mathrm{The}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rope}\:\mathrm{is}\:\mathrm{negligible}.\:\mathrm{In} \\ $$$$\left(\mathrm{a}\right),\:\mathrm{the}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{lifted}\:\mathrm{up}\:\mathrm{by}\:\mathrm{attaching} \\ $$$$\mathrm{a}\:\mathrm{mass}\:\left(\mathrm{2}{m}\right)\:\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rope}. \\ $$$$\mathrm{In}\:\left(\mathrm{b}\right),\:{m}\:\mathrm{is}\:\mathrm{lifted}\:\mathrm{up}\:\mathrm{by}\:\mathrm{pulling}\:\mathrm{the} \\ $$$$\mathrm{other}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rope}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\mathrm{downward}\:\mathrm{force}\:{F}\:=\:\mathrm{2}{mg}.\:\mathrm{In}\:\mathrm{which} \\ $$$$\mathrm{case},\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:{m}\:\mathrm{is}\:\mathrm{more}? \\ $$

Question Number 18320 Answers: 1 Comments: 0

In a triangle ABC with fixed base BC, the vertex A moves such that cos B + cos C = 4 sin^2 (A/2) . If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C respectively, then (1) b + c = 4a (2) b + c = 2a (3) Locus of point A is an ellipse (4) Locus of point A is a pair of straight lines

$$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:{ABC}\:\mathrm{with}\:\mathrm{fixed}\:\mathrm{base}\:{BC}, \\ $$$$\mathrm{the}\:\mathrm{vertex}\:{A}\:\mathrm{moves}\:\mathrm{such}\:\mathrm{that}\:\mathrm{cos}\:{B}\:+ \\ $$$$\mathrm{cos}\:{C}\:=\:\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}\:.\:\mathrm{If}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{denote} \\ $$$$\mathrm{the}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathrm{opposite}\:\mathrm{to}\:\mathrm{the}\:\mathrm{angles}\:{A},\:{B}\:\mathrm{and}\:{C} \\ $$$$\mathrm{respectively},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{b}\:+\:{c}\:=\:\mathrm{4}{a} \\ $$$$\left(\mathrm{2}\right)\:{b}\:+\:{c}\:=\:\mathrm{2}{a} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Locus}\:\mathrm{of}\:\mathrm{point}\:{A}\:\mathrm{is}\:\mathrm{an}\:\mathrm{ellipse} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Locus}\:\mathrm{of}\:\mathrm{point}\:{A}\:\mathrm{is}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{straight} \\ $$$$\mathrm{lines} \\ $$

Question Number 18274 Answers: 1 Comments: 0

A balloon moves up vertically such that if a stone is projected with a horizontal velocity u relative to balloon, the stone always hits the ground at a fixed point at a distance ((2u^2 )/g) horizontally away from it. Find the height of the balloon as a function of time.

$$\mathrm{A}\:\mathrm{balloon}\:\mathrm{moves}\:\mathrm{up}\:\mathrm{vertically}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{if}\:\mathrm{a}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{velocity}\:{u}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{balloon}, \\ $$$$\mathrm{the}\:\mathrm{stone}\:\mathrm{always}\:\mathrm{hits}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{fixed}\:\mathrm{point}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\frac{\mathrm{2}{u}^{\mathrm{2}} }{{g}}\:\mathrm{horizontally} \\ $$$$\mathrm{away}\:\mathrm{from}\:\mathrm{it}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{balloon}\:\mathrm{as}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\mathrm{time}. \\ $$

Question Number 18271 Answers: 0 Comments: 3

There are two parallel planes, each inclined to the horizontal at an angle θ. A particle is projected from a point mid way between the foot of the two planes so that it grazes one of the planes and strikes the other at right angle. Find the angle of projection of the projectile.

$$\mathrm{There}\:\mathrm{are}\:\mathrm{two}\:\mathrm{parallel}\:\mathrm{planes},\:\mathrm{each} \\ $$$$\mathrm{inclined}\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\theta. \\ $$$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{mid} \\ $$$$\mathrm{way}\:\mathrm{between}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{planes} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{it}\:\mathrm{grazes}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{planes}\:\mathrm{and} \\ $$$$\mathrm{strikes}\:\mathrm{the}\:\mathrm{other}\:\mathrm{at}\:\mathrm{right}\:\mathrm{angle}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{projection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{projectile}. \\ $$

Question Number 18269 Answers: 0 Comments: 3

The flow velocity of a river increases linearly with the distance (r) from its bank and has its maximum value v_0 in the middle of the river. The velocity near the bank is zero. A boat which can move with speed u in still water moves in the river in such a way that it is always perpendicular to the flow of current. Find (i) The distance along the bank through which boat is carried away by the flow current, when the boat crosses the river. (ii) The equation of trajectory for the coordinate system shown. Assume that the swimmer starts from origin.

Question Number 18268 Answers: 1 Comments: 0

A balloon starts rising from the surface of earth. The ascension rate is constant and is equal to v_0 . Due to wind the balloon gathers horizontal velocity component v_x = ay, where a is a positive constant and y is the height of ascent. Find (i) The horizontal drift of the balloon x(y), (ii) The total, tangential and normal accelerations of the balloon.

$$\mathrm{A}\:\mathrm{balloon}\:\mathrm{starts}\:\mathrm{rising}\:\mathrm{from}\:\mathrm{the}\:\mathrm{surface} \\ $$$$\mathrm{of}\:\mathrm{earth}.\:\mathrm{The}\:\mathrm{ascension}\:\mathrm{rate}\:\mathrm{is}\:\mathrm{constant} \\ $$$$\mathrm{and}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:{v}_{\mathrm{0}} .\:\mathrm{Due}\:\mathrm{to}\:\mathrm{wind}\:\mathrm{the} \\ $$$$\mathrm{balloon}\:\mathrm{gathers}\:\mathrm{horizontal}\:\mathrm{velocity} \\ $$$$\mathrm{component}\:{v}_{{x}} \:=\:{ay},\:\mathrm{where}\:{a}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{constant}\:\mathrm{and}\:{y}\:\mathrm{is}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{ascent}. \\ $$$$\mathrm{Find} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{The}\:\mathrm{horizontal}\:\mathrm{drift}\:\mathrm{of}\:\mathrm{the}\:\mathrm{balloon} \\ $$$${x}\left({y}\right), \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{total},\:\mathrm{tangential}\:\mathrm{and}\:\mathrm{normal} \\ $$$$\mathrm{accelerations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{balloon}. \\ $$

Question Number 19654 Answers: 0 Comments: 3

Two swimmers leave point A on one bank of the river to reach point B lying right across the other bank. One of them crosses the river along the straight line AB while the other swims at right angle to the stream and then walks the distance that he has been carried away by the stream to get to point B. What was the velocity v of his walking if both swimmers reached the destination simultaneously? (The stream velocity v_0 = 2 km/h and the velocity v′ of each swimmer with respect to still water is 2.5 km/h).

Question Number 18265 Answers: 0 Comments: 7

A particle is projected at an angle 60° with speed 10(√3) m/s from the point A as shown in the figure. At the same time the wedge is made to move with speed 10(√3) m/s toward right as shown in figure. Find the time after which particle will strike the wedge.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{60}° \\ $$$$\mathrm{with}\:\mathrm{speed}\:\mathrm{10}\sqrt{\mathrm{3}}\:\mathrm{m}/\mathrm{s}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:{A} \\ $$$$\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}.\:\mathrm{At}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{time}\:\mathrm{the}\:\mathrm{wedge}\:\mathrm{is}\:\mathrm{made}\:\mathrm{to}\:\mathrm{move}\:\mathrm{with} \\ $$$$\mathrm{speed}\:\mathrm{10}\sqrt{\mathrm{3}}\:\mathrm{m}/\mathrm{s}\:\mathrm{toward}\:\mathrm{right}\:\mathrm{as}\:\mathrm{shown} \\ $$$$\mathrm{in}\:\mathrm{figure}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{time}\:\mathrm{after}\:\mathrm{which} \\ $$$$\mathrm{particle}\:\mathrm{will}\:\mathrm{strike}\:\mathrm{the}\:\mathrm{wedge}. \\ $$

Question Number 18264 Answers: 1 Comments: 0

A sky diver of mass m drops out with an initial velocity v_0 = 0. Find the law by which the sky diver′s speed varies before the parachute is opened if the drag is proportional to the sky diver′s speed. Also solve the problem when the sky diver′s initial velocity has horizontal component v_0 and vertical component zero.

$$\mathrm{A}\:\mathrm{sky}\:\mathrm{diver}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{drops}\:\mathrm{out}\:\mathrm{with} \\ $$$$\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity}\:{v}_{\mathrm{0}} \:=\:\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{law} \\ $$$$\mathrm{by}\:\mathrm{which}\:\mathrm{the}\:\mathrm{sky}\:\mathrm{diver}'\mathrm{s}\:\mathrm{speed}\:\mathrm{varies} \\ $$$$\mathrm{before}\:\mathrm{the}\:\mathrm{parachute}\:\mathrm{is}\:\mathrm{opened}\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{drag}\:\mathrm{is}\:\mathrm{proportional}\:\mathrm{to}\:\mathrm{the}\:\mathrm{sky}\:\mathrm{diver}'\mathrm{s} \\ $$$$\mathrm{speed}.\:\mathrm{Also}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{sky}\:\mathrm{diver}'\mathrm{s}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{has}\:\mathrm{horizontal} \\ $$$$\mathrm{component}\:{v}_{\mathrm{0}} \:\mathrm{and}\:\mathrm{vertical}\:\mathrm{component} \\ $$$$\mathrm{zero}. \\ $$

Question Number 18262 Answers: 1 Comments: 2

A point P is located above an inclined plane. It is possible to reach the plane by sliding under gravity down a straight frictionless wire joining to some point P ′ on the plane. How should P ′ be chosen so as to minimize the time taken?

$$\mathrm{A}\:\mathrm{point}\:{P}\:\mathrm{is}\:\mathrm{located}\:\mathrm{above}\:\mathrm{an}\:\mathrm{inclined} \\ $$$$\mathrm{plane}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{by}\:\mathrm{sliding}\:\mathrm{under}\:\mathrm{gravity}\:\mathrm{down}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{frictionless}\:\mathrm{wire}\:\mathrm{joining}\:\mathrm{to}\:\mathrm{some}\:\mathrm{point} \\ $$$${P}\:'\:\mathrm{on}\:\mathrm{the}\:\mathrm{plane}.\:\mathrm{How}\:\mathrm{should}\:{P}\:'\:\mathrm{be} \\ $$$$\mathrm{chosen}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to}\:\mathrm{minimize}\:\mathrm{the}\:\mathrm{time} \\ $$$$\mathrm{taken}? \\ $$

Question Number 18206 Answers: 0 Comments: 0

1+((1×3)/6)+((1×3×5)/(6×8))+...∞

$$\mathrm{1}+\frac{\mathrm{1}×\mathrm{3}}{\mathrm{6}}+\frac{\mathrm{1}×\mathrm{3}×\mathrm{5}}{\mathrm{6}×\mathrm{8}}+...\infty \\ $$

Question Number 20973 Answers: 1 Comments: 1

A small solid spherical ball of high density is dropped in a viscous liquid. Its journey in the liquid is best described in the following figure by the curve

$$\mathrm{A}\:\mathrm{small}\:\mathrm{solid}\:\mathrm{spherical}\:\mathrm{ball}\:\mathrm{of}\:\mathrm{high} \\ $$$$\mathrm{density}\:\mathrm{is}\:\mathrm{dropped}\:\mathrm{in}\:\mathrm{a}\:\mathrm{viscous}\:\mathrm{liquid}. \\ $$$$\mathrm{Its}\:\mathrm{journey}\:\mathrm{in}\:\mathrm{the}\:\mathrm{liquid}\:\mathrm{is}\:\mathrm{best}\:\mathrm{described} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:\mathrm{figure}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve} \\ $$

Question Number 20972 Answers: 0 Comments: 0

Boyle temperature is given by (1) T_B = (a/(Rb^2 )) (2) T_B = (a/(Rb)) (3) T_B = (a/(27b^2 )) (4) T_B = (b/(aR))

$$\mathrm{Boyle}\:\mathrm{temperature}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{T}_{\mathrm{B}} \:=\:\frac{\mathrm{a}}{\mathrm{Rb}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:\mathrm{T}_{\mathrm{B}} \:=\:\frac{\mathrm{a}}{\mathrm{Rb}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{T}_{\mathrm{B}} \:=\:\frac{\mathrm{a}}{\mathrm{27b}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\mathrm{T}_{\mathrm{B}} \:=\:\frac{\mathrm{b}}{\mathrm{aR}} \\ $$

Question Number 20976 Answers: 0 Comments: 0

What would be the percentage composition by volume of a mixture of CO and CH_4 , whose 10.5 mL requires 9 mL oxygen for complete combustion?

$$\mathrm{What}\:\mathrm{would}\:\mathrm{be}\:\mathrm{the}\:\mathrm{percentage} \\ $$$$\mathrm{composition}\:\mathrm{by}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{a}\:\mathrm{mixture}\:\mathrm{of} \\ $$$$\mathrm{CO}\:\mathrm{and}\:\mathrm{CH}_{\mathrm{4}} ,\:\mathrm{whose}\:\mathrm{10}.\mathrm{5}\:\mathrm{mL}\:\mathrm{requires} \\ $$$$\mathrm{9}\:\mathrm{mL}\:\mathrm{oxygen}\:\mathrm{for}\:\mathrm{complete}\:\mathrm{combustion}? \\ $$

Question Number 18202 Answers: 1 Comments: 0

An open vessel at 27°C is heated until (3/5) parts of the air in it has been expelled. Assuming that the volume of the vessel remains constant, find the temperature to which the vessel has been heated.

$$\mathrm{An}\:\mathrm{open}\:\mathrm{vessel}\:\mathrm{at}\:\mathrm{27}°\mathrm{C}\:\mathrm{is}\:\mathrm{heated}\:\mathrm{until} \\ $$$$\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{parts}\:\mathrm{of}\:\mathrm{the}\:\mathrm{air}\:\mathrm{in}\:\mathrm{it}\:\mathrm{has}\:\mathrm{been}\:\mathrm{expelled}. \\ $$$$\mathrm{Assuming}\:\mathrm{that}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vessel} \\ $$$$\mathrm{remains}\:\mathrm{constant},\:\mathrm{find}\:\mathrm{the}\:\mathrm{temperature} \\ $$$$\mathrm{to}\:\mathrm{which}\:\mathrm{the}\:\mathrm{vessel}\:\mathrm{has}\:\mathrm{been}\:\mathrm{heated}. \\ $$

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