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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?

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

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.

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

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