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Question Number 20786 Answers: 2 Comments: 0

Two particles A and B start from the same position along the circular path of radius 0.5 m with a speed v_A = 1 ms^(−1) and v_B = 1.2 ms^(−1) in opposite direction. Determine the time before they collide.

$Two particles A and B start from the$ $same position along the circular path of$ $radius 0.5 m with a speed v_{A} = 1 {ms}^{- 1}$ $and v_{B} = 1.2 {ms}^{- 1} in opposite direction .$ $Determine the time before they collide .$

Question Number 20785 Answers: 1 Comments: 0

If in a ΔABC, tan(A/2), tan(B/2) and tan(C/2) are in HP, then the minimum value of cot(B/2) is greater than (1) 2(√3) (2) ((√3)/2) (3) (√3) (4) (1/(√3))

$If in a Δ A B C, \tan \frac{A}{2}, \tan \frac{B}{2} and \tan \frac{C}{2}$ $are in HP, then the minimum value of$ $\cot \frac{B}{2} is greater than$ $(1) 2 \sqrt{3}$ $(2) \frac{\sqrt{3}}{2}$ $(3) \sqrt{3}$ $(4) \frac{1}{\sqrt{3}}$

Question Number 20777 Answers: 0 Comments: 6

In the figure shown, mass ′m′ is placed on the inclined surface of a wedge of mass M. All the surfaces are smooth. Find the acceleration of the wedge.

$I n t h e f i g u r e s h o w n, m a s s^{'} m^{'} i s$ $p l a c e d o n t h e i n c l i n e d s u r f a c e o f a$ $w e d g e o f m a s s M . A l l t h e s u r f a c e s$ $a r e s m o o t h . F i n d t h e a c c e l e r a t i o n o f$ $t h e w e d g e .$

Question Number 20764 Answers: 1 Comments: 0

A small bead is slipped on a horizontal rod of length l. The rod starts moving with a horizontal acceleration a in a direction making an angle α with the length of the rod. Assuming that initially the bead is in the middle of the rod, find the time elapsed before the bead leaves the rod. Coefficient of friction between the bead and the rod is μ. (Neglect gravity).

$A s m a l l b e a d i s s l i p p e d o n a h o r i z o n t a l$ $r o d o f l e n g t h l . T h e r o d s t a r t s m o v i n g$ $w i t h a h o r i z o n t a l a c c e l e r a t i o n a i n a$ $d i r e c t i o n m a k i n g a n a n g l e α w i t h t h e$ $l e n g t h o f t h e r o d . A s s u m i n g t h a t$ $i n i t i a l l y t h e b e a d i s i n t h e m i d d l e o f$ $t h e r o d, f i n d t h e t i m e e l a p s e d b e f o r e$ $t h e b e a d l e a v e s t h e r o d . C o e f f i c i e n t o f$ $f r i c t i o n b e t w e e n t h e b e a d a n d t h e r o d$ $i s μ . (N e g l e c t g r a v i t y) .$

Question Number 20745 Answers: 1 Comments: 3

Consider a disc rotating in the horizontal plane with a constant angular speed ω about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the Figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed (1/8) rotation, (ii) their range is less than half the disc radius, and (iii) ω remains constant throughout. Then (a) P lands in the shaded region and Q in the unshaded region (b) P lands in the unshaded region and Q in the shaded region (c) Both P and Q land in the unshaded region (d) Both P and Q land in the shaded region

$Consider a disc rotating in the horizontal$ $plane with a constant angular speed ω$ $about its centre O . The disc has a$ $shaded region on one side of the diameter$ $and an unshaded region on the other$ $side as shown in the Figure . When the$ $disc is in the orientation as shown, two$ $pebbles P and Q are simultaneously$ $projected at an angle towards R . The$ $velocity of projection is in the y - z$ $plane and is same for both pebbles with$ $respect to the disc . Assume that (i) they$ $land back on the disc before the disc has$ $completed \frac{1}{8} rotation, (i i) their range$ $is less than half the disc radius, and$ $(i i i) ω remains constant throughout .$ $Then$ $(a) P lands in the shaded region and Q$ $in the unshaded region$ $(b) P lands in the unshaded region and$ $Q in the shaded region$ $(c) Both P and Q land in the unshaded$ $region$ $(d) Both P and Q land in the shaded$ $region$

Question Number 20744 Answers: 1 Comments: 0

Question Number 20739 Answers: 1 Comments: 0

If ∣z_1 ∣ = 2, ∣z_2 ∣ = 3, ∣z_3 ∣ = 4 and ∣2z_1 + 3z_2 + 4z_3 ∣ = 4, then the expression ∣8z_2 z_3 + 27z_3 z_1 + 64z_1 z_2 ∣ equals

$If ∣ z_{1} ∣ = 2, ∣ z_{2} ∣ = 3, ∣ z_{3} ∣ = 4 and$ $∣ 2 z_{1} + 3 z_{2} + 4 z_{3} ∣ = 4, then the expression$ $∣ 8 z_{2} z_{3} + 27 z_{3} z_{1} + 64 z_{1} z_{2} ∣ equals$

Question Number 20733 Answers: 1 Comments: 0

Find the volume of the solid of revolution obtained by revolving area bounded by x = 4 + 6y − 2y^2 , x = −4, x = 0 about the y−axis

$Find the volume of the solid of revolution$ $obtained by revolving area bounded by$ $x = 4 + 6 y - 2 y^{2}, x = - 4, x = 0 about$ $the y - axis$

Question Number 20724 Answers: 0 Comments: 3

A sphere is rolling without slipping on a fixed horizontal plane surface. In the Figure, A is a point of contact, B is the centre of the sphere and C is its topmost point. Then (a) v_C ^→ − v_A ^→ = 2(v_B ^→ − v_C ^→ ) (b) v_C ^→ − v_B ^→ = v_B ^→ − v_A ^→ (c) ∣v_C ^→ − v_A ^→ ∣ = 2∣v_B ^→ − v_C ^→ ∣ (d) ∣v_C ^→ − v_A ^→ ∣ = 4∣v_B ^→ ∣

$A sphere is rolling without slipping on$ $a fixed horizontal plane surface . In the$ $Figure, A is a point of contact, B is the$ $centre of the sphere and C is its topmost$ $point . Then$ $(a) {\vec{v}}_{C} - {\vec{v}}_{A} = 2 ({\vec{v}}_{B} - {\vec{v}}_{C})$ $(b) {\vec{v}}_{C} - {\vec{v}}_{B} = {\vec{v}}_{B} - {\vec{v}}_{A}$ $(c) ∣ {\vec{v}}_{C} - {\vec{v}}_{A} ∣ = 2 ∣ {\vec{v}}_{B} - {\vec{v}}_{C} ∣$ $(d) ∣ {\vec{v}}_{C} - {\vec{v}}_{A} ∣ = 4 ∣ {\vec{v}}_{B} ∣$

Question Number 20723 Answers: 0 Comments: 0

∫(√x) .sinx.dx

$\int \sqrt{x} . s i n x . d x$

Question Number 20721 Answers: 1 Comments: 0

tan xtan z=3 tan ytan z=6 x+y+z = π Solve for x, y, and z.

$\tan x \tan z = 3$ $\tan y \tan z = 6$ $x + y + z = π$ $S o l v e f o r x, y, a n d z .$

Question Number 20705 Answers: 1 Comments: 0

Question Number 20704 Answers: 1 Comments: 2

4cos θcos (((2π)/3)+θ)cos (((4π)/3)+θ)=cos 3θ

$4 \cos θ \cos (\frac{2 π}{3} + θ) \cos (\frac{4 π}{3} + θ) = \cos 3 θ$

Question Number 20703 Answers: 1 Comments: 1

((sinα+sin 3α+sin5α)/(cos α+cos 3α+cos 5α))=tan 3α

$\frac{\sin α + \sin 3 α + \sin 5 α}{\cos α + \cos 3 α + \cos 5 α} = \tan 3 α$

Question Number 20702 Answers: 1 Comments: 0

tan 20°tan 40°tan 80°=(√3)

$\tan 20 ° \tan 40 ° \tan 80 ° = \sqrt{3}$

Question Number 20693 Answers: 1 Comments: 0

Integers 1, 2, 3, ...., n, where n > 2, are written on a board. Two numbers m, k such that 1 < m < n, 1 < k < n are removed and the average of the remaining numbers is found to be 17. What is the maximum sum of the two removed numbers?

$Integers 1, 2, 3, . . . ., n, where n > 2, are$ $written on a board . Two numbers m, k$ $such that 1 < m < n, 1 < k < n are$ $removed and the average of the$ $remaining numbers is found to be 17 .$ $What is the maximum sum of the two$ $removed numbers ?$

Question Number 20761 Answers: 0 Comments: 5

A 5 kg block B is suspended from a cord attached to a 40 kg cart A. Find the accelerations of both the block and cart. (All surfaces are frictionless) (g = 10 m/s^2 )

$A 5 k g b l o c k B i s s u s p e n d e d f r o m a$ $c o r d a t t a c h e d t o a 40 k g c a r t A . F i n d$ $t h e a c c e l e r a t i o n s o f b o t h t h e b l o c k a n d$ $c a r t . (A l l s u r f a c e s a r e f r i c t i o n l e s s)$ $(g = 10 m / s^{2})$

Question Number 20686 Answers: 2 Comments: 1

∫(tanx)^(1/3) dx

$\int {(t a n x)}^{1 / 3} d x$

Question Number 20684 Answers: 1 Comments: 0

If the equation x^2 + β^2 = 1 − 2βx and x^2 + α^2 = 1 − 2αx have one and only one root in common, then ∣α − β∣ is equal to

$I f t h e e q u a t i o n x^{2} + β^{2} = 1 - 2 β x a n d$ $x^{2} + α^{2} = 1 - 2 α x h a v e o n e a n d o n l y$ $o n e r o o t i n c o m m o n, t h e n ∣ α - β ∣ i s$ $e q u a l t o$

Question Number 20671 Answers: 1 Comments: 0

The total number of positive integral solution(s) of the inequation ((x^2 (3x − 4)^3 (x − 2)^4 )/((x − 5)^5 (2x − 7)^6 )) ≤ 0 is/are

$T h e t o t a l n u m b e r o f p o s i t i v e i n t e g r a l$ $s o l u t i o n (s) o f t h e i n e q u a t i o n$ $\frac{x^{2} {(3 x - 4)}^{3} {(x - 2)}^{4}}{{(x - 5)}^{5} {(2 x - 7)}^{6}} ⩽ 0 i s / a r e$

Question Number 20666 Answers: 1 Comments: 0

∫cot^4 xdx

$\int \cot^{4} x d x$

Question Number 20670 Answers: 1 Comments: 0

For the equation 3x^2 + px + 3 = 0, find the value(s) of p if one root is (i) square of the other (ii) fourth power of the other.

$F o r t h e e q u a t i o n 3 x^{2} + p x + 3 = 0,$ $f i n d t h e v a l u e (s) o f p i f o n e r o o t i s$ $(i) s q u a r e o f t h e o t h e r$ $(i i) f o u r t h p o w e r o f t h e o t h e r .$

Question Number 20653 Answers: 2 Comments: 0

Length of interval of range of function f(θ) = cos^2 θ − 6 sin θ cos θ + 3 sin^2 θ + 2 is (1) 8 (2) −8 (3) (√(10)) (4) 2(√(10))

$Length of interval of range of function$ $f (θ) = \cos^{2} θ - 6 \sin θ \cos θ + 3 \sin^{2} θ + 2$ $is$ $(1) 8$ $(2) - 8$ $(3) \sqrt{10}$ $(4) 2 \sqrt{10}$

Question Number 20652 Answers: 0 Comments: 4

Suppose x is a positive real number such that {x}, [x] and x are in a geometric progression. Find the least positive integer n such that x^n > 100. (Here [x] denotes the integer part of x and {x} = x − [x].)

$Suppose x is a positive real number$ $such that {x}, [x] and x are in a$ $geometric progression . Find the least$ $positive integer n such that x^{n} > 100 .$ $(Here [x] denotes the integer part of x$ $and {x} = x - [x] .)$

Question Number 20648 Answers: 1 Comments: 0

If (F/(sin (A − a))) = (W/(sin A)), prove that tan A = ((W sin a)/(W cos a − F))

$If \frac{F}{\sin (A - a)} = \frac{W}{\sin A}, prove that$ $\tan A = \frac{W \sin a}{W \cos a - F}$

Question Number 20647 Answers: 1 Comments: 0

prove it, tan (α+(π/3))+tan (α−(π/3))=((4sin 2α)/(1−4sin^2 α))

$p r o v e i t,$ $\tan (α + \frac{π}{3}) + \tan (α - \frac{π}{3}) = \frac{4 \sin 2 α}{1 - 4 \sin^{2} α}$

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