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Question Number 20927 Answers: 1 Comments: 0

(a+b)×(a+b)

$$\left({a}+{b}\right)×\left({a}+{b}\right) \\ $$

Question Number 20916 Answers: 1 Comments: 0

A body starts rotating about a stationary axis with an angular acceleration b = 2t rad/s^2 . How soon after the beginning of rotation will the total acceleration vector of an arbitrary point on the body forms an angle of 60° with its velocity vector? (1) (2(√3))^(1/3) s (2) (2(√3))^(1/2) s (3) (2(√3)) s (4) (2(√3))^2 s

$$\mathrm{A}\:\mathrm{body}\:\mathrm{starts}\:\mathrm{rotating}\:\mathrm{about}\:\mathrm{a} \\ $$$$\mathrm{stationary}\:\mathrm{axis}\:\mathrm{with}\:\mathrm{an}\:\mathrm{angular} \\ $$$$\mathrm{acceleration}\:{b}\:=\:\mathrm{2}{t}\:\mathrm{rad}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{How}\:\mathrm{soon} \\ $$$$\mathrm{after}\:\mathrm{the}\:\mathrm{beginning}\:\mathrm{of}\:\mathrm{rotation}\:\mathrm{will}\:\mathrm{the} \\ $$$$\mathrm{total}\:\mathrm{acceleration}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{an}\:\mathrm{arbitrary} \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{body}\:\mathrm{forms}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{60}° \\ $$$$\mathrm{with}\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{vector}? \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}} \:\mathrm{s} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{2}} \:\mathrm{s} \\ $$$$\left(\mathrm{3}\right)\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)\:\mathrm{s} \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\mathrm{s} \\ $$

Question Number 20915 Answers: 1 Comments: 0

Two shells are fired from a canon with speed u each, at angles of α and β respectively with the horizontal. The time interval between the shots is t. They collide in mid air after time T from the first shot. Which of the following conditions must be satisfied? (a) α > β (b) T cos α = (T − t) cos β (c) (T − t) cos α = T cos β (d) u sin α T − (1/2) g T^2 = u sin β (T − t) − (1/2) g (T − t)^2

$$\mathrm{Two}\:\mathrm{shells}\:\mathrm{are}\:\mathrm{fired}\:\mathrm{from}\:\mathrm{a}\:\mathrm{canon}\:\mathrm{with}\:\mathrm{speed}\:\mathrm{u}\:\mathrm{each},\:\mathrm{at} \\ $$$$\mathrm{angles}\:\mathrm{of}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{respectively}\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{The} \\ $$$$\mathrm{time}\:\mathrm{interval}\:\mathrm{between}\:\mathrm{the}\:\mathrm{shots}\:\mathrm{is}\:{t}.\:\mathrm{They}\:\mathrm{collide}\:\mathrm{in}\:\mathrm{mid} \\ $$$$\mathrm{air}\:\mathrm{after}\:\mathrm{time}\:{T}\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{shot}.\:\mathrm{Which}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{conditions}\:\mathrm{must}\:\mathrm{be}\:\mathrm{satisfied}? \\ $$$$\left({a}\right)\:\alpha\:>\:\beta \\ $$$$\left({b}\right)\:{T}\:\mathrm{cos}\:\alpha\:=\:\left({T}\:−\:{t}\right)\:\mathrm{cos}\:\beta \\ $$$$\left({c}\right)\:\left({T}\:−\:{t}\right)\:\mathrm{cos}\:\alpha\:=\:{T}\:\mathrm{cos}\:\beta \\ $$$$\left({d}\right)\:{u}\:\mathrm{sin}\:\alpha\:{T}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:{g}\:{T}^{\mathrm{2}} \:=\:{u}\:\mathrm{sin}\:\beta\:\left({T}\:−\:{t}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:{g}\:\left({T}\:−\:{t}\right)^{\mathrm{2}} \\ $$

Question Number 20914 Answers: 1 Comments: 0

The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has (1) Only purely imaginary roots (2) All real roots (3) Two real and two purely imaginary roots (4) Neither real nor purely imaginary roots

$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equation}\:{p}\left({x}\right)\:=\:\mathrm{0}\:\mathrm{with} \\ $$$$\mathrm{real}\:\mathrm{coefficients}\:\mathrm{has}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{equation}\:{p}\left({p}\left({x}\right)\right)\:=\:\mathrm{0} \\ $$$$\mathrm{has} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Only}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{roots} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{All}\:\mathrm{real}\:\mathrm{roots} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Two}\:\mathrm{real}\:\mathrm{and}\:\mathrm{two}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Neither}\:\mathrm{real}\:\mathrm{nor}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots} \\ $$

Question Number 20908 Answers: 1 Comments: 0

integrate with respect to x ∫(((2x+1)/(x^2 +4x+8)))dx

$${integrate}\:{with}\:{respect}\:{to}\:{x}\: \\ $$$$\int\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}}\right){dx} \\ $$

Question Number 20907 Answers: 1 Comments: 0

How many zeroes (0) there in 1×2×3×4..........99×100

$${How}\:{many}\:{zeroes}\:\left(\mathrm{0}\right)\:{there}\:{in}\:\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}..........\mathrm{99}×\mathrm{100}\: \\ $$

Question Number 20905 Answers: 1 Comments: 1

∫e^x^2 dx

$$\int{e}^{{x}^{\mathrm{2}} } {dx} \\ $$

Question Number 20903 Answers: 1 Comments: 0

Question Number 20891 Answers: 0 Comments: 11

The Figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed ω/2. The ring and disc are separated by frictionless ball bearing. The system is in the x-z plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle 30° with the horizontal. Then with respect to the horizontal surface (a) The point O has a linear velocity 3Rωi^∧ (b) The point P has a linear velocity ((11)/4)Rωi^∧ + ((√3)/4)Rωk^∧

$$\mathrm{The}\:\mathrm{Figure}\:\mathrm{shows}\:\mathrm{a}\:\mathrm{system}\:\mathrm{consisting} \\ $$$$\mathrm{of}\:\left({i}\right)\:\mathrm{a}\:\mathrm{ring}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{radius}\:\mathrm{3}{R}\:\mathrm{rolling} \\ $$$$\mathrm{clockwise}\:\mathrm{without}\:\mathrm{slipping}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{with}\:\mathrm{angular}\:\mathrm{speed} \\ $$$$\omega\:\mathrm{and}\:\left({ii}\right)\:\mathrm{an}\:\mathrm{inner}\:\mathrm{disc}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{2}{R} \\ $$$$\mathrm{rotating}\:\mathrm{anti}-\mathrm{clockwise}\:\mathrm{with}\:\mathrm{angular} \\ $$$$\mathrm{speed}\:\omega/\mathrm{2}.\:\mathrm{The}\:\mathrm{ring}\:\mathrm{and}\:\mathrm{disc}\:\mathrm{are} \\ $$$$\mathrm{separated}\:\mathrm{by}\:\mathrm{frictionless}\:\mathrm{ball}\:\mathrm{bearing}. \\ $$$$\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:{x}-{z}\:\mathrm{plane}.\:\mathrm{The} \\ $$$$\mathrm{point}\:{P}\:\mathrm{on}\:\mathrm{the}\:\mathrm{inner}\:\mathrm{disc}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance} \\ $$$${R}\:\mathrm{from}\:\mathrm{the}\:\mathrm{origin},\:\mathrm{where}\:{OP}\:\mathrm{makes}\:\mathrm{an} \\ $$$$\mathrm{angle}\:\mathrm{30}°\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{Then} \\ $$$$\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{surface} \\ $$$$\left({a}\right)\:\mathrm{The}\:\mathrm{point}\:{O}\:\mathrm{has}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{velocity} \\ $$$$\mathrm{3}{R}\omega\overset{\wedge} {{i}} \\ $$$$\left({b}\right)\:\mathrm{The}\:\mathrm{point}\:{P}\:\mathrm{has}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{velocity} \\ $$$$\frac{\mathrm{11}}{\mathrm{4}}{R}\omega\overset{\wedge} {{i}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{R}\omega\overset{\wedge} {{k}} \\ $$

Question Number 20886 Answers: 1 Comments: 0

if sin x=msin y so proof that tan (1/2)(x−y)=((m−1)/(m+1))tan (1/2)(x+y)

$${if}\:\mathrm{sin}\:{x}={m}\mathrm{sin}\:{y} \\ $$$${so}\:{proof}\:{that} \\ $$$$\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)=\frac{{m}−\mathrm{1}}{{m}+\mathrm{1}}\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}\right) \\ $$

Question Number 20885 Answers: 0 Comments: 0

if (θ−ϕ)subtle and sin θ+sin ϕ= (cos ϕ−cos θ)(√3) so proof sin 3θ+sin 3ϕ=0

$${if}\:\left(\theta−\varphi\right){subtle}\:{and}\:\:\:\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi= \\ $$$$\left(\mathrm{cos}\:\varphi−\mathrm{cos}\:\theta\right)\sqrt{\mathrm{3}} \\ $$$${so}\:{proof}\:\mathrm{sin}\:\mathrm{3}\theta+\mathrm{sin}\:\mathrm{3}\varphi=\mathrm{0} \\ $$

Question Number 20884 Answers: 1 Comments: 0

2cos (π/3)cos ((9π)/(13))+cos ((3π)/(13))+cos ((5π)/(13))=0

$$\mathrm{2cos}\:\frac{\pi}{\mathrm{3}}\mathrm{cos}\:\frac{\mathrm{9}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{13}}=\mathrm{0} \\ $$

Question Number 20882 Answers: 1 Comments: 0

Question Number 20881 Answers: 2 Comments: 0

Question Number 20876 Answers: 0 Comments: 1

Question Number 20873 Answers: 1 Comments: 0

∫_1 ^5 (e^x /x^2 ) dx

$$\int_{\mathrm{1}} ^{\mathrm{5}} \frac{{e}^{{x}} }{{x}^{\mathrm{2}} }\:{dx} \\ $$

Question Number 20872 Answers: 1 Comments: 0

if y=[xtan^(−1) x]−[(1/2)ln(1+x^2 )] show that (1+x^2 )y^(′′) =1

$${if}\:\:{y}=\left[{xtan}^{−\mathrm{1}} {x}\right]−\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right] \\ $$$${show}\:{that}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{''} =\mathrm{1} \\ $$

Question Number 20871 Answers: 1 Comments: 0

Question Number 20870 Answers: 0 Comments: 0

Question Number 20869 Answers: 0 Comments: 0

Question Number 20868 Answers: 0 Comments: 1

C×(A+B)=C×A+C×B prove it.

$$\boldsymbol{{C}}×\left(\boldsymbol{{A}}+\boldsymbol{{B}}\right)=\boldsymbol{{C}}×\boldsymbol{{A}}+\boldsymbol{{C}}×\boldsymbol{{B}}\: \\ $$$${prove}\:{it}. \\ $$

Question Number 20857 Answers: 0 Comments: 0

∫ (√(sin x)) + (√(cos x)) dx

$$\int\:\sqrt{\mathrm{sin}\:{x}}\:+\:\sqrt{\mathrm{cos}\:{x}}\:{dx} \\ $$

Question Number 20853 Answers: 0 Comments: 0

$$ \\ $$

Question Number 20851 Answers: 2 Comments: 0

if (x+y)^(m+n) =x^m y^n ,show that (dy/dx)=(y/x)

$${if}\:\left({x}+{y}\right)^{{m}+{n}} ={x}^{{m}} {y}^{{n}} ,{show}\:{that}\:\frac{{dy}}{{dx}}=\frac{{y}}{{x}} \\ $$

Question Number 20850 Answers: 1 Comments: 0

given that y=log(√((1+cos^2 x)/(1−e^(2x) ))),find (dy/dx)

$${given}\:{that}\:{y}={log}\sqrt{\frac{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{\mathrm{1}−{e}^{\mathrm{2}{x}} }},{find} \\ $$$$\frac{{dy}}{{dx}} \\ $$

Question Number 20849 Answers: 1 Comments: 0

The time period,T of pendulum of length l is given by T=2Π(√(l/g))where l amd g are constant.Find the approximate percentage increase in time T, when the length of pendulum increases by 4%

$${The}\:{time}\:{period},{T}\:\:{of}\:{pendulum}\:{of} \\ $$$${length}\:{l}\:{is}\:{given}\:{by}\:{T}=\mathrm{2}\Pi\sqrt{\frac{{l}}{{g}}}{where} \\ $$$$\:{l}\:{amd}\:{g}\:{are}\:{constant}.{Find}\:{the}\: \\ $$$${approximate}\:{percentage}\:{increase}\: \\ $$$${in}\:{time}\:{T},\:{when}\:{the}\:{length}\:{of}\: \\ $$$${pendulum}\:{increases}\:{by}\:\mathrm{4\%} \\ $$

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