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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.

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

Question Number 20977 Answers: 0 Comments: 1

Imtegrate ∫e^(−ax^2 +bx+c) dx for a>0. It′s just for fun. If you have questions leave a comment. I′ll do my best to answer them.

$${Imtegrate}\:\int{e}^{−{ax}^{\mathrm{2}} +{bx}+{c}} {dx}\:{for}\:{a}>\mathrm{0}. \\ $$$${It}'{s}\:{just}\:{for}\:{fun}.\:{If}\:{you}\:{have}\:{questions} \\ $$$${leave}\:{a}\:{comment}.\:{I}'{ll}\:{do}\:{my}\:{best}\:{to}\:{answer}\:{them}. \\ $$

Question Number 18199 Answers: 0 Comments: 0

What is the equivalent weight of KH(IO_3 )_2 as an oxidant in presence of 4 (N) HCl when ICl becomes the reduced form? (K = 39, I = 127)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{equivalent}\:\mathrm{weight}\:\mathrm{of} \\ $$$$\mathrm{KH}\left(\mathrm{IO}_{\mathrm{3}} \right)_{\mathrm{2}} \:\mathrm{as}\:\mathrm{an}\:\mathrm{oxidant}\:\mathrm{in}\:\mathrm{presence}\:\mathrm{of} \\ $$$$\mathrm{4}\:\left(\mathrm{N}\right)\:\mathrm{HCl}\:\mathrm{when}\:\mathrm{ICl}\:\mathrm{becomes}\:\mathrm{the} \\ $$$$\mathrm{reduced}\:\mathrm{form}?\:\left(\mathrm{K}\:=\:\mathrm{39},\:\mathrm{I}\:=\:\mathrm{127}\right) \\ $$

Question Number 18198 Answers: 1 Comments: 0

Mixture X = 0.02 mol of [Co(NH_3 )_5 SO_4 ]Br and 0.02 mol of [Co(NH_3 )_5 Br]SO_4 was prepared in 2 litre of solution 1 litre of mixture X + excess AgNO_3 → Y 1 litre of mixture X + excess BaCl_2 → Z Number of moles of Y and Z are

$$\mathrm{Mixture}\:\mathrm{X}\:=\:\mathrm{0}.\mathrm{02}\:\mathrm{mol}\:\mathrm{of} \\ $$$$\left[\mathrm{Co}\left(\mathrm{NH}_{\mathrm{3}} \right)_{\mathrm{5}} \mathrm{SO}_{\mathrm{4}} \right]\mathrm{Br}\:\mathrm{and}\:\mathrm{0}.\mathrm{02}\:\mathrm{mol}\:\mathrm{of} \\ $$$$\left[\mathrm{Co}\left(\mathrm{NH}_{\mathrm{3}} \right)_{\mathrm{5}} \mathrm{Br}\right]\mathrm{SO}_{\mathrm{4}} \:\mathrm{was}\:\mathrm{prepared}\:\mathrm{in}\:\mathrm{2} \\ $$$$\mathrm{litre}\:\mathrm{of}\:\mathrm{solution} \\ $$$$\mathrm{1}\:\mathrm{litre}\:\mathrm{of}\:\mathrm{mixture}\:\mathrm{X}\:+\:\mathrm{excess}\:\mathrm{AgNO}_{\mathrm{3}} \:\rightarrow\:\mathrm{Y} \\ $$$$\mathrm{1}\:\mathrm{litre}\:\mathrm{of}\:\mathrm{mixture}\:\mathrm{X}\:+\:\mathrm{excess}\:\mathrm{BaCl}_{\mathrm{2}} \:\rightarrow\:\mathrm{Z} \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{moles}\:\mathrm{of}\:\mathrm{Y}\:\mathrm{and}\:\mathrm{Z}\:\mathrm{are} \\ $$

Question Number 18197 Answers: 1 Comments: 0

Rearrange the following (I to IV) in the the order of increasing masses. I. 1 molecule of oxygen II. 1 atom of nitrogen III. 10^(10) g molecular weight of oxygen IV. 10^(−18) g atomic weight of copper

$$\mathrm{Rearrange}\:\mathrm{the}\:\mathrm{following}\:\left(\mathrm{I}\:\mathrm{to}\:\mathrm{IV}\right)\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{increasing}\:\mathrm{masses}. \\ $$$$\mathrm{I}.\:\mathrm{1}\:\mathrm{molecule}\:\mathrm{of}\:\mathrm{oxygen} \\ $$$$\mathrm{II}.\:\mathrm{1}\:\mathrm{atom}\:\mathrm{of}\:\mathrm{nitrogen} \\ $$$$\mathrm{III}.\:\mathrm{10}^{\mathrm{10}} \:\mathrm{g}\:\mathrm{molecular}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{oxygen} \\ $$$$\mathrm{IV}.\:\mathrm{10}^{−\mathrm{18}} \:\mathrm{g}\:\mathrm{atomic}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{copper} \\ $$

Question Number 18186 Answers: 2 Comments: 0

3.92 g of ferrous ammonium sulphate are dissolved in 100 ml of water. 20 ml of this solution requires 18 ml of potassium permanganate during titration for complete oxidation. The weight of KMnO_4 present in one litre of the solution is

$$\mathrm{3}.\mathrm{92}\:\mathrm{g}\:\mathrm{of}\:\mathrm{ferrous}\:\mathrm{ammonium}\:\mathrm{sulphate} \\ $$$$\mathrm{are}\:\mathrm{dissolved}\:\mathrm{in}\:\mathrm{100}\:\mathrm{ml}\:\mathrm{of}\:\mathrm{water}.\:\mathrm{20}\:\mathrm{ml} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{solution}\:\mathrm{requires}\:\mathrm{18}\:\mathrm{ml}\:\mathrm{of} \\ $$$$\mathrm{potassium}\:\mathrm{permanganate}\:\mathrm{during} \\ $$$$\mathrm{titration}\:\mathrm{for}\:\mathrm{complete}\:\mathrm{oxidation}.\:\mathrm{The} \\ $$$$\mathrm{weight}\:\mathrm{of}\:\mathrm{KMnO}_{\mathrm{4}} \:\mathrm{present}\:\mathrm{in}\:\mathrm{one}\:\mathrm{litre} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is} \\ $$

Question Number 18185 Answers: 0 Comments: 2

Is this true or false? acosA + bcosB + ccosC = ((abc)/(2R^2 ))

$$\mathrm{Is}\:\mathrm{this}\:\mathrm{true}\:\mathrm{or}\:\mathrm{false}? \\ $$$${a}\mathrm{cos}{A}\:+\:{b}\mathrm{cos}{B}\:+\:{c}\mathrm{cos}{C}\:=\:\frac{{abc}}{\mathrm{2}{R}^{\mathrm{2}} } \\ $$

Question Number 18184 Answers: 1 Comments: 2

(m+2)sinθ + (2m−1)cosθ = 2m+1, if (1) tanθ = (3/4) (2) tanθ = (4/3) (3) tanθ = ((2m)/(m^2 − 1)) (4) tanθ = ((2m)/(m^2 + 1))

$$\left({m}+\mathrm{2}\right)\mathrm{sin}\theta\:+\:\left(\mathrm{2}{m}−\mathrm{1}\right)\mathrm{cos}\theta\:=\:\mathrm{2}{m}+\mathrm{1},\:\mathrm{if} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{2}{m}}{{m}^{\mathrm{2}} \:−\:\mathrm{1}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{2}{m}}{{m}^{\mathrm{2}} \:+\:\mathrm{1}} \\ $$

Question Number 18257 Answers: 0 Comments: 2

Question Number 18261 Answers: 1 Comments: 0

What is the maximum angle to the horizontal at which a stone can be thrown and always be moving away from the thrower?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{angle}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{at}\:\mathrm{which}\:\mathrm{a}\:\mathrm{stone}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{thrown}\:\mathrm{and}\:\mathrm{always}\:\mathrm{be}\:\mathrm{moving}\:\mathrm{away} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{thrower}? \\ $$

Question Number 18258 Answers: 0 Comments: 0

Question Number 18168 Answers: 1 Comments: 0

If the coefficient of the middle term in the expansion of the (1+x)^(2n+2) is p and the coefficients of middle terms in the expansion of (1+x)^(2n+1) are q and r, then

$$\mathrm{If}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{term}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\mathrm{the}\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}+\mathrm{2}} \:\mathrm{is}\:{p}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{coefficients}\:\mathrm{of}\:\mathrm{middle}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}+\mathrm{1}} \:\mathrm{are}\:{q}\:\mathrm{and}\:{r},\:\mathrm{then} \\ $$

Question Number 18153 Answers: 0 Comments: 1

Question Number 18152 Answers: 0 Comments: 0

Question Number 18149 Answers: 1 Comments: 2

Question Number 18146 Answers: 0 Comments: 1

Question Number 18142 Answers: 1 Comments: 1

An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0, the plank starts moving along the x- direction with an acceleration 1.5 ms^(−2) . At the same instant a stone is projected from the origin with a velocity u^→ as shown. A stationary person on the ground observe the stone hitting the object during its downward motion at an angle of 45° with the horizontal. Take g = 10 m/s^2 and consider all motions in the x-y plane. 1. The time after which the stone hits the object is 2. The initial velocity (u^→ ) of the particle is

Question Number 18140 Answers: 1 Comments: 0

From a tower of height H, a particle is thrown vertically upward with speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is (1) 2gH = n^2 u^2 (2) gH = (n − 2)^2 u^2 (3) 2gH = nu^2 (n − 2) (4) gH = (n − 2)u^2

$$\mathrm{From}\:\mathrm{a}\:\mathrm{tower}\:\mathrm{of}\:\mathrm{height}\:{H},\:\mathrm{a}\:\mathrm{particle}\:\mathrm{is} \\ $$$$\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with}\:\mathrm{speed} \\ $$$${u}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle},\:\mathrm{to} \\ $$$$\mathrm{hit}\:\mathrm{the}\:\mathrm{ground},\:\mathrm{is}\:{n}\:\mathrm{times}\:\mathrm{that}\:\mathrm{taken}\:\mathrm{by} \\ $$$$\mathrm{it}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{highest}\:\mathrm{point}\:\mathrm{of}\:\mathrm{its}\:\mathrm{path}. \\ $$$$\mathrm{The}\:\mathrm{relation}\:\mathrm{between}\:{H},\:{u}\:\mathrm{and}\:{n}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}{gH}\:=\:{n}^{\mathrm{2}} {u}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{gH}\:=\:\left({n}\:−\:\mathrm{2}\right)^{\mathrm{2}} {u}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2}{gH}\:=\:{nu}^{\mathrm{2}} \left({n}\:−\:\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)\:{gH}\:=\:\left({n}\:−\:\mathrm{2}\right){u}^{\mathrm{2}} \\ $$

Question Number 18135 Answers: 0 Comments: 6

Let x be the LCM of 3^(2002) − 1 and 3^(2002) + 1. Find the last digit of x.

$$\mathrm{Let}\:{x}\:\mathrm{be}\:\mathrm{the}\:\mathrm{LCM}\:\mathrm{of}\:\mathrm{3}^{\mathrm{2002}} \:−\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{3}^{\mathrm{2002}} \:+\:\mathrm{1}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{of}\:{x}. \\ $$

Question Number 18131 Answers: 0 Comments: 3

A stone is projected from a point on the ground in such a direction so as to hit a bird on the top of a telegraph post of height h, and then attain a height 2h above the ground. If, at an instant of projection, the bird were to fly away horizontal with a uniform speed, find the ratio of the horizontal velocities of the bird and the stone, if the stone still hits the bird.

$$\mathrm{A}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{direction}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to}\:\mathrm{hit}\:\mathrm{a} \\ $$$$\mathrm{bird}\:\mathrm{on}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a}\:\mathrm{telegraph}\:\mathrm{post}\:\mathrm{of} \\ $$$$\mathrm{height}\:{h},\:\mathrm{and}\:\mathrm{then}\:\mathrm{attain}\:\mathrm{a}\:\mathrm{height}\:\mathrm{2}{h} \\ $$$$\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{If},\:\mathrm{at}\:\mathrm{an}\:\mathrm{instant}\:\mathrm{of} \\ $$$$\mathrm{projection},\:\mathrm{the}\:\mathrm{bird}\:\mathrm{were}\:\mathrm{to}\:\mathrm{fly}\:\mathrm{away} \\ $$$$\mathrm{horizontal}\:\mathrm{with}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{speed},\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{velocities}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{bird}\:\mathrm{and}\:\mathrm{the}\:\mathrm{stone},\:\mathrm{if}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{still} \\ $$$$\mathrm{hits}\:\mathrm{the}\:\mathrm{bird}. \\ $$

Question Number 18133 Answers: 1 Comments: 0

A 4×4×4 wooden cube is painted so that one pair of opposite faces is blue, one pair green and one pair red. The cube is now sliced into 64 cubes of side 1 unit each. (i) How many of the smaller cubes have no painted face? (ii) How many of the smaller cubes have exactly one painted face? (iii) How many of the smaller cubes have exactly two painted face? (iv) How many of the smaller cubes have exactly three painted face? (v) How many of the smaller cubes have exactly one face painted blue and one face painted green?

$$\mathrm{A}\:\mathrm{4}×\mathrm{4}×\mathrm{4}\:\mathrm{wooden}\:\mathrm{cube}\:\mathrm{is}\:\mathrm{painted}\:\mathrm{so} \\ $$$$\mathrm{that}\:\mathrm{one}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{opposite}\:\mathrm{faces}\:\mathrm{is}\:\mathrm{blue}, \\ $$$$\mathrm{one}\:\mathrm{pair}\:\mathrm{green}\:\mathrm{and}\:\mathrm{one}\:\mathrm{pair}\:\mathrm{red}.\:\mathrm{The} \\ $$$$\mathrm{cube}\:\mathrm{is}\:\mathrm{now}\:\mathrm{sliced}\:\mathrm{into}\:\mathrm{64}\:\mathrm{cubes}\:\mathrm{of}\:\mathrm{side} \\ $$$$\mathrm{1}\:\mathrm{unit}\:\mathrm{each}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{cubes}\:\mathrm{have} \\ $$$$\mathrm{no}\:\mathrm{painted}\:\mathrm{face}? \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{cubes}\:\mathrm{have} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{painted}\:\mathrm{face}? \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{cubes}\:\mathrm{have} \\ $$$$\mathrm{exactly}\:\mathrm{two}\:\mathrm{painted}\:\mathrm{face}? \\ $$$$\left(\mathrm{iv}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{cubes}\:\mathrm{have} \\ $$$$\mathrm{exactly}\:\mathrm{three}\:\mathrm{painted}\:\mathrm{face}? \\ $$$$\left(\mathrm{v}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{cubes}\:\mathrm{have} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{face}\:\mathrm{painted}\:\mathrm{blue}\:\mathrm{and}\:\mathrm{one} \\ $$$$\mathrm{face}\:\mathrm{painted}\:\mathrm{green}? \\ $$

Question Number 18167 Answers: 0 Comments: 3

Question Number 18123 Answers: 0 Comments: 2

Question Number 18114 Answers: 0 Comments: 0

(√(((1+cosθ)(1+cosθ))/(1−cosθ)(1+cosθ)))) ⇒(√(((1+cosθ)^2 )/(1−cos^2 θ))) recall that:cos^2 θ−1=sin^2 θ ((1+cosθ)/(sin^2 θ)) (1/(sinθ))+((cosθ)/(sinθ)) =cosecθ+cotθ

$$\sqrt{\frac{\left(\mathrm{1}+\mathrm{cos}\theta\right)\left(\mathrm{1}+\mathrm{cos}\theta\right)}{\left.\mathrm{1}−\mathrm{cos}\theta\right)\left(\mathrm{1}+\mathrm{cos}\theta\right)}} \\ $$$$\Rightarrow\sqrt{\frac{\left(\mathrm{1}+\mathrm{cos}\theta\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta}} \\ $$$$\mathrm{recall}\:\mathrm{that}:\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{1}=\mathrm{sin}^{\mathrm{2}} \theta \\ $$$$\frac{\mathrm{1}+\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\theta}+\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta} \\ $$$$=\mathrm{cosec}\theta+\mathrm{cot}\theta \\ $$

Question Number 18111 Answers: 1 Comments: 3

Question Number 18109 Answers: 0 Comments: 4

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