Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 1819

Question Number 18278    Answers: 2   Comments: 1

Question Number 18279    Answers: 0   Comments: 0

In an 1800 m race, P beats Q by 50 seconds. In the same race, Q beats R by 40 seconds. If P beats R by 450 m, by what distance does P beat Q ?(in m)

$$\mathrm{In}\:\mathrm{an}\:\mathrm{1800}\:\mathrm{m}\:\mathrm{race},\:\mathrm{P}\:\mathrm{beats}\:\mathrm{Q}\:\mathrm{by}\:\mathrm{50}\: \\ $$$$\mathrm{seconds}.\:\mathrm{In}\:\mathrm{the}\:\mathrm{same}\:\mathrm{race},\:\mathrm{Q}\:\mathrm{beats}\:\mathrm{R}\:\mathrm{by} \\ $$$$\mathrm{40}\:\mathrm{seconds}.\:\mathrm{If}\:\mathrm{P}\:\mathrm{beats}\:\mathrm{R}\:\mathrm{by}\:\mathrm{450}\:\mathrm{m},\:\mathrm{by}\: \\ $$$$\mathrm{what}\:\mathrm{distance}\:\mathrm{does}\:\mathrm{P}\:\mathrm{beat}\:\mathrm{Q}\:?\left(\mathrm{in}\:\mathrm{m}\right) \\ $$

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.

$$\mathrm{The}\:\mathrm{flow}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{river}\:\mathrm{increases} \\ $$$$\mathrm{linearly}\:\mathrm{with}\:\mathrm{the}\:\mathrm{distance}\:\left({r}\right)\:\mathrm{from}\:\mathrm{its} \\ $$$$\mathrm{bank}\:\mathrm{and}\:\mathrm{has}\:\mathrm{its}\:\mathrm{maximum}\:\mathrm{value}\:{v}_{\mathrm{0}} \:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{middle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{river}.\:\mathrm{The}\:\mathrm{velocity} \\ $$$$\mathrm{near}\:\mathrm{the}\:\mathrm{bank}\:\mathrm{is}\:\mathrm{zero}.\:\mathrm{A}\:\mathrm{boat}\:\mathrm{which}\:\mathrm{can} \\ $$$$\mathrm{move}\:\mathrm{with}\:\mathrm{speed}\:{u}\:\mathrm{in}\:\mathrm{still}\:\mathrm{water}\:\mathrm{moves} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{river}\:\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{always}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{the}\:\mathrm{flow}\:\mathrm{of} \\ $$$$\mathrm{current}.\:\mathrm{Find} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{The}\:\mathrm{distance}\:\mathrm{along}\:\mathrm{the}\:\mathrm{bank}\:\mathrm{through} \\ $$$$\mathrm{which}\:\mathrm{boat}\:\mathrm{is}\:\mathrm{carried}\:\mathrm{away}\:\mathrm{by}\:\mathrm{the}\:\mathrm{flow} \\ $$$$\mathrm{current},\:\mathrm{when}\:\mathrm{the}\:\mathrm{boat}\:\mathrm{crosses}\:\mathrm{the} \\ $$$$\mathrm{river}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{trajectory}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{coordinate}\:\mathrm{system}\:\mathrm{shown}.\:\mathrm{Assume} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{swimmer}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{origin}. \\ $$

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).

$$\mathrm{Two}\:\mathrm{swimmers}\:\mathrm{leave}\:\mathrm{point}\:{A}\:\mathrm{on}\:\mathrm{one} \\ $$$$\mathrm{bank}\:\mathrm{of}\:\mathrm{the}\:\mathrm{river}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{point}\:{B}\:\mathrm{lying} \\ $$$$\mathrm{right}\:\mathrm{across}\:\mathrm{the}\:\mathrm{other}\:\mathrm{bank}.\:\mathrm{One}\:\mathrm{of} \\ $$$$\mathrm{them}\:\mathrm{crosses}\:\mathrm{the}\:\mathrm{river}\:\mathrm{along}\:\mathrm{the}\:\mathrm{straight} \\ $$$$\mathrm{line}\:{AB}\:\mathrm{while}\:\mathrm{the}\:\mathrm{other}\:\mathrm{swims}\:\mathrm{at}\:\mathrm{right} \\ $$$$\mathrm{angle}\:\mathrm{to}\:\mathrm{the}\:\mathrm{stream}\:\mathrm{and}\:\mathrm{then}\:\mathrm{walks}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{that}\:\mathrm{he}\:\mathrm{has}\:\mathrm{been}\:\mathrm{carried}\:\mathrm{away} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{stream}\:\mathrm{to}\:\mathrm{get}\:\mathrm{to}\:\mathrm{point}\:{B}.\:\mathrm{What} \\ $$$$\mathrm{was}\:\mathrm{the}\:\mathrm{velocity}\:{v}\:\mathrm{of}\:\mathrm{his}\:\mathrm{walking}\:\mathrm{if}\:\mathrm{both} \\ $$$$\mathrm{swimmers}\:\mathrm{reached}\:\mathrm{the}\:\mathrm{destination} \\ $$$$\mathrm{simultaneously}?\:\left(\mathrm{The}\:\mathrm{stream}\:\mathrm{velocity}\right. \\ $$$${v}_{\mathrm{0}} \:=\:\mathrm{2}\:\mathrm{km}/\mathrm{h}\:\mathrm{and}\:\mathrm{the}\:\mathrm{velocity}\:{v}'\:\mathrm{of}\:\mathrm{each} \\ $$$$\mathrm{swimmer}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{still}\:\mathrm{water}\:\mathrm{is} \\ $$$$\left.\mathrm{2}.\mathrm{5}\:\mathrm{km}/\mathrm{h}\right). \\ $$

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 18247    Answers: 0   Comments: 0

(((x+1)(x−1))/(x^2 +2x+3))

$$\frac{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}} \\ $$

Question Number 18243    Answers: 0   Comments: 2

Question Number 18236    Answers: 0   Comments: 0

∫ (dx/(1 − sin x + cos x))

$$\int\:\frac{{dx}}{\mathrm{1}\:−\:\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}} \\ $$

Question Number 18233    Answers: 1   Comments: 2

If y = (√(x^2 + (√(x^2 + (√(x^2 + (√(...)))))))) find ∫(y + (√y)) dx

$$\mathrm{If}\:{y}\:=\:\sqrt{{x}^{\mathrm{2}} \:+\:\sqrt{{x}^{\mathrm{2}} \:+\:\sqrt{{x}^{\mathrm{2}} \:+\:\sqrt{...}}}} \\ $$$$\mathrm{find}\:\int\left({y}\:+\:\sqrt{{y}}\right)\:{dx} \\ $$

Question Number 18232    Answers: 1   Comments: 0

If f(x) = (√(x^2 (√(3(√(x^2 (√(3(√)...)))))))) find ∫ f(x) dx

$$\mathrm{If}\:{f}\left({x}\right)\:=\:\sqrt{{x}^{\mathrm{2}} \sqrt{\mathrm{3}\sqrt{{x}^{\mathrm{2}} \sqrt{\mathrm{3}\sqrt{}...}}}} \\ $$$$\mathrm{find}\:\int\:{f}\left({x}\right)\:{dx} \\ $$

Question Number 18224    Answers: 1   Comments: 0

∫cos2x ln(1+tanx)dx

$$\int\mathrm{cos2x}\:\mathrm{ln}\left(\mathrm{1}+\mathrm{tanx}\right)\mathrm{dx} \\ $$

Question Number 18223    Answers: 1   Comments: 0

prove that ∫_0 ^( π) ((x tanx)/(tanx+secx))dx=(π/2)(π−2)

$$\mathrm{prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\:\pi} \frac{\mathrm{x}\:\mathrm{tanx}}{\mathrm{tanx}+\mathrm{secx}}\mathrm{dx}=\frac{\pi}{\mathrm{2}}\left(\pi−\mathrm{2}\right) \\ $$

Question Number 18222    Answers: 0   Comments: 1

show that ∫_0 ^(π/4) ((x sinx)/(1+cos^2 x))dx=(π/4)

$$\mathrm{show}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{x}\:\mathrm{sinx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}=\frac{\pi}{\mathrm{4}} \\ $$

Question Number 18214    Answers: 1   Comments: 0

The speeds of Daniel and Robert are in the ratio of 3:4.In a race of 300m Daniel has a start of 90m. Daniel won by?

$$\mathrm{The}\:\mathrm{speeds}\:\mathrm{of}\:\mathrm{Daniel}\:\mathrm{and}\:\mathrm{Robert} \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{3}:\mathrm{4}.\mathrm{In}\:\mathrm{a}\:\mathrm{race}\:\mathrm{of} \\ $$$$\mathrm{300m}\:\mathrm{Daniel}\:\mathrm{has}\:\mathrm{a}\:\mathrm{start}\:\mathrm{of}\:\mathrm{90m}.\: \\ $$$$\mathrm{Daniel}\:\mathrm{won}\:\mathrm{by}? \\ $$

Question Number 18213    Answers: 1   Comments: 0

The greatest number that will divide 82 ,111 and 140 leaving the same remainder in each case is........

$$\mathrm{The}\:\mathrm{greatest}\:\mathrm{number}\:\mathrm{that}\:\mathrm{will} \\ $$$$\mathrm{divide}\:\mathrm{82}\:,\mathrm{111}\:\mathrm{and}\:\mathrm{140}\:\mathrm{leaving}\:\mathrm{the}\: \\ $$$$\mathrm{same}\:\mathrm{remainder}\:\mathrm{in}\:\mathrm{each}\:\mathrm{case}\:\mathrm{is}........ \\ $$

Question Number 18209    Answers: 2   Comments: 1

Question Number 18207    Answers: 1   Comments: 0

If ((3+5+7+...+n terms)/(5+8+11+...+10 terms)) = 7, then the value of n is

$$\mathrm{If}\:\frac{\mathrm{3}+\mathrm{5}+\mathrm{7}+...+{n}\:\mathrm{terms}}{\mathrm{5}+\mathrm{8}+\mathrm{11}+...+\mathrm{10}\:\mathrm{terms}}\:=\:\mathrm{7},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:{n}\:\mathrm{is} \\ $$

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?

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

  Pg 1814      Pg 1815      Pg 1816      Pg 1817      Pg 1818      Pg 1819      Pg 1820      Pg 1821      Pg 1822      Pg 1823   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com