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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
Two equal forces act at a point. The square of their resultant is ` 3 ` times their product, Find the angle between them. |
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Answer» Correct Answer - `60^(@)` |
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| 552. |
Vector `vec(A)` is 2`cm` long and is `60^(@)` above the x-axis in the first quadrant. Vector `vec(B)` is `2 cm` long and is `60^(@)` below the x-axis in the fourth quadrant. The sum `vec(A)+vec(B)` is a vector of magnitudes |
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Answer» `vecR=vecA+vecB` `vecR=2cos 60^(@)hati+2sin60^(@)hatj+2cos 60^(@)hati-2sin60^(@)hatj` `vecR=4cos 60^(@)hati` `therefore R=2cm`, along x-axis |
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| 553. |
Two forces, each equal to F, act as shown in (figure) Their resultant is A. `F//2`B. `F`C. `sqrt(3) F`D. `sqrt(5)F` |
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Answer» Correct Answer - B Note that the angle between two forces is `120^(@)` and not `160^(@)` `R^(2)=F^(2) +F^(2)+2F^(2) cos 120^(@)` `=2F^(2)+2F^(2) (-1/2) F^(2)` or `R=F` |
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| 554. |
A river is flowing from west to east at a speed of 5 metres per minute.A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction.A. due northB. `30^(@)` east of northC. `30^(@)` west of northD. `60^(@)` east of north |
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Answer» Correct Answer - A For shortest time, swimmer swin at `90^(@)` `t_("min")=(d)/(v_(s)sin theta)` ` theta= 90` `sin theta=1` |
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| 555. |
Vector `vec(A)` is 2`cm` long and is `60^(@)` above the x-axis in the first quadrant. Vactor `vec(B)` is `2 cm` long and is `60^(@)` below the x-axis in the fourth quadrant. The sum `vec(A)+vec(B)` is a vector of magnitudesA. 2`cm` along positive y-axisB. 2`cm` along positive x-axisC. 2`cm` along negative y-axisD. 2`cm` along negative x-axis |
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Answer» Correct Answer - B Here the angle between two vectors of equal magnitude is 120. So resultant has the same magnitude as either of the given vectors. Moreover, it is mid way between the two vectors, i.e., it is also x-axis. |
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| 556. |
Resultant of two vetors A and B is given by `|R|={|A|-|B|}.`angle between A and B will beA. `90^(@)`B. `180^(@)`C. `0^(@)`D. none of these |
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Answer» Correct Answer - B (d ) Resultant of two vector A and B `R={|A|-|B|}` `then theta=180^(@)` |
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| 557. |
Which is a vector quantity?A. Angular momentumB. WorkC. Potential energyD. Electirc current |
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Answer» Correct Answer - A (a) Angular momentum is a vector quantity. |
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| 558. |
Which one is a vector quantity?A. WorkB. momentumC. timeD. Speed |
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Answer» Correct Answer - B (b) Momentum is the product of mass and velocity.mass is a scalar quantity and velocity is a vector quantity. So momentum is a vector quantity. |
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| 559. |
Which is a vector quantity?A. WorkB. powerC. TorqueD. Gravitational constant |
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Answer» Correct Answer - C (c ) Torque is defined as `t=rxxF`.So ,it is a vector quantity. |
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| 560. |
If `vecA=(1)/sqrt(2)cos theta hati+(1)/sqrt(2)sin thetahatj`, what will be the unit vector perpendiuclar to `vecA`.A. `cos theta hati+sin thetahatj`B. `-cos theta hati+sin thetahatj`C. `(cos thetahati+sin thetahatj)/sqrt(2)`D. `sin thetahati-cos theta hatj` |
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Answer» Correct Answer - D `vecAxxvecB=Abhatn, hatB=(vecB)/(B)` |
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| 561. |
Which of the following is the unit vector perpendicular to `vec(A)` and `vec(B)`?A. `(hatAxxhatB)/(ABsintheta)`B. `(hatAxxhatB)/(ABcostheta)`C. `(AxxB)/(ABsintheta)`D. `(AxxB)/(ABcostheta)` |
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Answer» Correct Answer - C |
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| 562. |
If `vec(a) = hati - 2hatj - 3hatk, vec(b) = 2 hati - hatj - hatk` and `vec(c) = hati +3 hatj - 2hatk` find `(vec(a) xx vec(b)) xx vec(c)` |
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Answer» Correct Answer - `-5 hati + 15 hatj +20 hatk` |
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| 563. |
`(hati+hatj)xx(hati-hatj)=`A. `-2hatk`B. `2hatk`C. zeroD. `2hati` |
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Answer» Correct Answer - A properties of vector product. |
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| 564. |
The unit vector parallel to the resultant of the vectors `vecA=4hati+3hatj+6hatk` and `vecB=-hati+3hatj-8hatk` isA. `(1)/(7)(3hati+6hatj-2hatk)`B. `(1)/(7)(3hati+6hatj+2hatk)`C. `(1)/(49)(3hati+6hatj-2hatk)`D. `(1)/(49)(3hati-6hatj+2hatk)` |
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Answer» Correct Answer - A `hatn=(vecA+vecB)/(|vecA+vecB|)`, |
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| 565. |
The unit vector parallel to the resultant of the vectors `vecA=4hati+3hatj+6hatk` and `vecB=-hati+3hatj-8hatk` isA. `1/7(3hati+6hatj-2hatk)`B. `1/7(3hati+6hatj+2hatk)`C. `1/49(3hati+6hatj-2hatk)`D. `1/49(3hati-6hatj+2hatk)` |
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Answer» Correct Answer - A |
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| 566. |
A boy walks uniformly along the sides of a rectangular park of size `400mxx300m`, starting from one corner to the other corner diagonally opposite. Which of the following statements is incorrect?A. He has travelled a distance of `700m`B. His displacement is `700m`C. His displacement is `500m`D. His velocity is unit throughout the walk |
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Answer» Correct Answer - B |
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| 567. |
A bird moves with velocity 20m/s in a direction making an angle of `60^(@)` with the eastern line and `60^(@)` with the vertical upward. Represent the velocity vector in rectangular form. |
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Answer» Let eastern line by taken as x-axis, northern as y-axis and vertical upward as z-axis Let the velocity `vecv` makes angle `alpha,beta` and `gamma` with x,y and z axis respectively, then `alpha=60^(@)`, `gamma=60^(@)` we have `cos^(2)alpha+cos^(2)beta+cos^(2)gamma=1` `cos^(2)60^(@)+cos^(2)beta+cos^(2)60^(@)=1` `cos^(2)beta=(1)/(2), cos beta=(1)/sqrt(2)` so `vecv=v cos alpha hati+v cos beta hatj+v cos gammahatk` `=20[(1)/(2)hati+(1)/sqrt(2)hati+(1)/(2)hatk]` `=10hati+10sqrt(2)hatj+10hatk` |
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| 568. |
A bird moves with velocity `20m s^(-1)` in a direction making an angle of `60^(@)` with vertical upward .Represent the velocity vector in a rectangular form. |
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Answer» Let the easten line be taken as x-axis ,northen as y-axis,and vertical upward as z-axis . Let the velocity`v` makes angles `alpha, beta`, and `gamma` with x-, y- and z-axis, respectively. Then `alpha=60^(@)` and `gamma=60^(@)`. `cos^(2)alpha+cos^(2)beta+cos^(2)gamma=1` We have `cos^(2)60^(@)+cos^(2)beta+cos^(2)60^(@)=1` or `cos beta=1/sqrt(2)` ` vec(v)=vcos alpha hat(i)+v cos beta hat (j)+v cos gamma hat(k)` `=20[1/2hat(i)+1/sqrt(2)hat(j)+1/2hat(k)]=10hat(i)+10sqrt(2)hat(j)+10hat(k)` |
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| 569. |
If `vec(A)= 2hat(i)+4hat(j)-5hat(k)` then the direction of cosins of the vector `vec(A)` areA. `2/(sqrt(45)),4/(sqrt(45)) and (-5)/(sqrt(45))`B. `1/(sqrt(45)),2/(sqrt(45)) and (3)/(sqrt(45))`C. `4/(sqrt(45)),0 and (4)/(sqrt(45))`D. `3/(sqrt(45)),2/(sqrt(45)) and (5)/(sqrt(45))` |
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Answer» Correct Answer - A `vec(A)= 2hat(i)+4hat(j)-5hat(k) :. |vec(A)|= sqrt((2)^(2)+(4)^(2)+(-5)^(2))=sqrt(45)` `:. cos alpha = 2/(sqrt(45)), cos beta=(4)/(sqrt(45)), cos gamma=(-5)/(sqrt(45))` |
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| 570. |
Given `vec(A)=5hat(i)+2hat(j)+4hat(k)`. Find (a) `|vec(A)|` and (b) the direction cosines of vector `vec(A)`. |
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Answer» (a) As `vec(A)=5hat(i)+2hat(j)+4hat(k)rArr|vec(A)|=sqrt(25+4+16)=sqrt(45)` (b) cos `alpha=l=x/r=5/sqrt(45),cos beta=m=y/r=2/sqrt(45),` `cos gamma=n=z/r=4/sqrt(45)` |
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| 571. |
If `vec(A) = 3 hat(i) + 6 hat(j) - 2hat(k)` , the directions of cosines of the vector `vec(A)` areA. `(3)/(7) ,(6)/(7) ,(2)/(7)`B. `(3)/(7) ,(6)/(7) , (-2)/(7)`C. `(6)/(7) ,(2)/(7) ,(3)/(7)`D. `(2)/(7) ,(3)/(7) ,(6)/(7)` |
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Answer» Correct Answer - B If `vec (r ) = x hat(i) + y hat(j) + z hat(k) , r = sqrt(x^(2) + y^(2) + z^(2))` `alpha , beta , gamma` : angles made by `vec(r )` with coordinates axes `cos alpha = (x)/( r) , cos beta = (y) /( r ) , cos gamma = (z)/( r )` Direction cosines ,` cos alpha , cos beta , cos gamma` `vec(A) = 3 i + 6 j - 2k , |vec(A)|= A = sqrt((3)^(2) + (6)^(2) + (-2)^(2)) = 7` `cos alpha = (3)/(7) , cos beta = (6)/(7) , cos gamma = (-2)/(7)` |
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| 572. |
If a vector `vec(A)` make angles `alpha , beta ` and `gamma`, respectively , with the `X , Y` and `Z` axes , then `sin^(2) alpha + sin^(2) beta + sin^(2) gamma =`A. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - C `cos^(2) alpha + cos^(2) beta + cos^(2) gamma = 1` `1 - sin^(2) alpha + 1 - sin^(2) beta + 1 - sin^(2) gamma = 1` `sin^(2) alpha + sin^(2) beta + sin^(2) gamma = 2` |
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| 573. |
If `vec(A)` makes an angle `alpha, beta` and `gamma` from x,y and z axis respectively then `sin^(2)alpha+sin^(2) beta+sin^(2) gamma=`A. 3B. 2C. 1D. 0 |
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Answer» Correct Answer - B `" "cos^(2)alpha+ cos^(2) beta+ cos^(2) gamma=1` `" "(1-sin^(2)alpha)+(1-sin ^(2) beta)+(1- sin ^(2)gamma)=1` ` therefore" "sin^(2) alpha + sin ^(2) beta+sin ^(2) gamma=2` |
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| 574. |
Three vectors `veca,vecb` and `vecc` satisfy the relation `veca.vecb=0` and `veca.vecc=0`. The vector `veca` is parallel toA. `vecB`B. `vecC`C. `vecB . vecC`D. `vecB xx vecC` |
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Answer» Correct Answer - D `vecA. vecB = 0" "and " " vecA. vecC=0` `rArr vecA` is `bot to vecB and vecC`. `So, vecB ` is parallel to `vecC` Then , `vecA` is parallel to `vecC.` |
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| 575. |
Given `A= 3hati+4hatj and B=6hati+8hatj ` which of the following statement is correct ?A. `AxxB=0`B. `(|A|)/(|B|)=(1)/(2)`C. `|A|=15`D. `A.|B|=48` |
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Answer» Correct Answer - A,B `(a,b)|A|=sqrt(9+16)=5 and |B|=sqrt(36+54)=10B=2AorB` is parallel to A ratio fo their coefficients are equal ,so they are parallel Or their cross product is zero. |
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| 576. |
Unit vector parallel to the resultant of vectors `A = 4hatj - 3hatj and B =8hatj+8hatj` will beA. `(24hati+5hatj)/(13)`B. `(12hati+5hatj)/(13)`C. `(6hati+5hatj)/(13)`D. None of these |
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Answer» Correct Answer - B `R = A+B = 12hati+5hatj` `R = |R| = sqrt((12)^2 + (5)^2) =13` `hatR = (R )/(R )`. |
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| 577. |
The angles which the vector `A=3hati + 6hatj+2hatk` makes with the co-ordinate axes areA. `cos^(-1).(3)/(7), cos^(-1),(6)/(7) and cos^(-1), (2)/(7)`B. `cos^(-1).(4)/(7), cos^(-1),(5)/(7) and cos^(-1),(3)/(7)`C. `cos^(-1).(3)/(7), cos^(-1),(4)/(7) and cos^(-1),(1)/(7)`D. None of these |
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Answer» Correct Answer - A `A = |A| = sqrt((3)^2 +(6)^2 +(2))^2` =7 `alpha = cos^(-1) ((A_x)/(A)) = cos ^(-1) ((3)/(7))` = angle of A with positivej x-axis. Similarly, `beta and lambda` angles. |
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| 578. |
The vector that must be added to the vector `hati-3hatj+2hatk` and `3hati+6hatj+7hatk` so that the resultant vector is a unit vector along the y-axis isA. `4hati +2hatj +5hatk `B. `-4hati +2hatj +5hatk `C. `3hati +4hatj +5hatk `D. Null vector |
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Answer» Correct Answer - B (b) `hatj=(hati-3hatj+2hatk)+(3hati+6hatj-7hatk)+C` `"Here",C=-4hati-2hatj+5hatk` |
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| 579. |
The vectors from origin to the points A and B are `vecA=3hati-6hatj+2hatk` and `vecB=2hati+hatj+2hatk` respectively. The are of triangle `OAB` beA. `5/2 sqrt(17)` sq. unitB. `2/5 sqrt(17)` sq. unitC. `3/5sqrt(17)` sq. unitD. `5/3sqrt(17)` sq. unit |
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Answer» Correct Answer - A |
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| 580. |
The vector that must be added to the vector `hati-3hatj+2hatk` and `3hati+6hatj+7hatk` so that the resultant vector is a unit vector along the y-axis isA. `4hati+2hatj+5hatk`B. `-4hati-2hatj+5hatk`C. `3hati+4hatj+5hatk`D. Null vector |
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Answer» Correct Answer - B |
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| 581. |
The angles which the vector `A=3hati + 6hatj+2hatk` makes with the co-ordinate axes areA. `cos^(-1)""(3)/(7)cos^(-1)""(6)/(7)and cos^(-1)""(2)/(7)`B. `cos^(-1)""(4)/(7)cos^(-1)""(5)/(7)and cos^(-1)""(3)/(7)`C. `cos^(-1)""(3)/(7)cos^(-1)""(4)/(7)and cos^(-1)""(1)/(7)`D. None of the above |
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Answer» Correct Answer - A (a) Vector`A=sqrt((3)^(2)+(6)^(2)+(2)^(2))=7` `alpha=cos^(-1)((3)/(7)),beta=cos^(-1)((6)/(7)),gamma=cos^(-1)((2)/(i))` |
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| 582. |
Let `veca=(1,1,-1), vecb=(5,-3,-3)` and `vecc=(3,-1,2)`. If `vecr` is collinear with `vecc` and has length `(|veca+vecb|)/(2)`, then `vecr` equalsA. `+-3vecc`B. `+-3/2vecc`C. `+-vecc`D. `+-2/3vecc` |
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Answer» Correct Answer - C We have, `vecr=lambdavecc` Given, `|vecr|=|lambda||vecr|` `therefore |6hati-2hatj-hatk|=2|lambda||3hati-hatj+2hatk|` `therefore sqrt(56) = 2|lambda|sqrt(14)` `therefore lambda=+-1` `therefore vecr=+-vecc` |
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| 583. |
The vectors `x hati + (x+1)hatj + (x+2)hatk, (x+3)hati+ (x+4)hatj + (x+5)hatk and (x+6)hati + (x+7)hatj+ (x+8)hatk` are coplanar if x is equal toA. 1B. -3C. 4D. 0 |
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Answer» Correct Answer - A::B::C::D ` x hati + (x + 1) hatj + (x +2) hatk , ( x+3) hati + (x +4) hatj + ( x+5) hatk and ( x+6) hati + (x+7) hatj+ ( x+8) hatk` are coplanar. We have determinant of their coefficients as `|{:(x,,x+1,,x+2),(x+3,,x+4,,x+5),(x+6,,x+7,,x+8):}|` Applying `C_2 to C_2 - C_1 and C_3 to C_3-C_1`, we have `|{:(x,,1,,2),(x+3,,1,,2),(x+6,,1,,2):}|=0` Here , `x in R`. |
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| 584. |
If `veca` and `vecb` are position vectors of A and B respectively, then the position vector of a point C in `vec(AB)` produced such that `vec(AC)` =2015 `vec(AB)` isA. `2014 veca-2015vecb`B. `2014vecb+2015veca`C. `2015vecb+2014veca`D. `2015vecb-2014veca` |
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Answer» Correct Answer - D `vec(AC) 2015 vec(AB)` `therefore vecc-veca=2015(vecb-veca)` `therefore vecc=2015 vec(b)-2014veca` |
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| 585. |
The unit vector bisecting `vec(OY)` and `vec(OZ)` isA. `(veci+vecj+veck)/sqrt(3)`B. `(veci-veck)/sqrt(2)`C. `(vecj+veck)/sqrt(2)`D. `(-vecj+veck)/sqrt(2)` |
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Answer» Correct Answer - C `hatj` and `hatk` are unit vectors along `Y-` and `Z-` axes, then unit vector bisecting `vec(OY)` and `vec(OZ)` is `(hati+hatk)/sqrt(2)` |
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| 586. |
If `vecx and vecy ` are two non-collinear vectors and a, b and c represent the sides of a `Delta ABC` satisfying `(a-b)vec x + (b-c)vecy+ (c-a)(vecxxx vecy) =0`, then `Delta ABC` is (where `vecx xx vecy` is perpendicular to the plane of `vecx and vecy`)A. an acute-angled triangleB. an obtuse-angled triangleC. a right-angled triangleD. a scalene triangle |
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Answer» Correct Answer - A As `vecx, vecy and vecx xx vecy` are non-collinear vectors, vectors are linearly independent. Hence, `" "a - b=0 = b -c =c -a ` or `a = b = c` Therefore, the triangle is equilateral. |
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| 587. |
A unit tangent vector at t=2 on the curve `x=t^(2)+2, y=4t-5` and `z=2t^(2)-6t` isA. `1/sqrt(3)(veci+vecj+veck)`B. `1/3(2veci+2vecj+veck)`C. `1/sqrt(6)(2veci+vecj+veck)`D. `1/3(veci+vecj+veck)` |
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Answer» Correct Answer - B `vecr = (vect^(2)+2)hati+(4t-5)hatj + (2t^(2)-6t)hatk` `rArr (dvecr)/(dt) = 2thati + 4hatj+(4t-6)hatk` `rArr (dvecr)/(dt)_(r=2) = 4hati+4hatj+2hatk` `rArr` Unit tangent vector at `t=2` is `1/3(2hati+2hatj+hatk)` |
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| 588. |
A uni-modular tangentvector on the curve `x=t^2+2,y=4t-5,z=2t^2-6t=2`isa. `1/3(2 hat i+2 hat j+ hat k)`b.`1/3( hat i- hat j- hat k)`c. `1/6(2 hat i+ hat j+ hat k)`d. `2/3( hat i+ hat j+ hat k)`A. `(1)/(3)(2hati+ 2hatj+hatk)`B. `(1)/(3) (hati-hatj-hatk)`C. `(1)/(6)(2hati + hatj + hatk)`D. `(2)/(3)(hati +hatj +hatk)` |
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Answer» Correct Answer - A The position vector of any point at `t` is `vecr = (2+ t^(2) ) hati + (4t -5) hatj + ( 2t^(2) -6) hatk` `rArr (dvecr)/( dt) = 2t hai + 4hatj + ( 4t -6) hatk` `rArr " "(dvecr)/(dt):|_(t=2) = 4hati +4hatj+2hatk` and `" "|(dvecr)/(dt)| :|_(t=2) = sqrt(16+16+4)=4` Hence, the required unit tangent vector at `t = 2` is `(1)/(3) ( 2hati + 2hatj +hatk)`. |
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| 589. |
Which of the following qauntities is dependent of the choice of orientation of coordinates axes?A. `vec(A)+vec(B)`B. `A_(x)+B_(y)`C. `|vec(A)+vec(B)|`D. Angle between A and B |
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Answer» Correct Answer - B A vector, its magnitude and the angle between two vectors do not depend on the choice of the orientation of the coordinates axes. So `vec(A)+vec(B),|vec(A)+vec(B)|`, angle between `vec(A) and vec(B)` are independent of the orientation of the coordinates axes. |
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| 590. |
Which of the following will not depend on orientation of frame of reference ?A. A scalarB. A vectorC. The magnitude of a vectorD. Component of a vector |
| Answer» Correct Answer - A::B::C | |
| 591. |
A force `vecF=(5hati+3hatj)` Newton is applied over a particle which displaces it from its origin to the point `vecr=(2hati-1hatj)` metres. The work done on the particle isA. `-7J`B. `+13J`C. `+7J`D. `+11J` |
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Answer» Correct Answer - C |
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| 592. |
The angle between two vectors `-2hati+3hatj+k` and `hati+2hatj-4hatk` isA. `0^(@)`B. `90^(@)`C. `180^(@)`D. None of the above |
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Answer» Correct Answer - B |
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| 593. |
The angle between the two vectors `-2hati+3hatj-hatk` and `hati+2hatj+4hatk` isA. `0^(@)`B. `90^(@)`C. `180^(@)`D. None of these |
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Answer» Correct Answer - B `theta cos^(-1) (vecA.vecB//AB),` Here `vecA. vecB = 0, so theta= 90^(@)` |
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| 594. |
The angle between two vectors `-2hati+3hatj+k` and `hati+2hatj-4hatk` isA. `45^(@)`B. `90^(@)`C. `30^(@)`D. `60^(@)` |
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Answer» Correct Answer - B `(b) Let A=-2hati+3hatj+hatk` `B=hati+2hatj-4hatk` `thereforeA.B =(-2hati+3hatj+hatk).(hati+hatj-4hatk)=-2+6-4=0` Since dot product of two vector is zero so vrctor will be perpendicular each other |
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| 595. |
Out of the following set of forces, the rsultant of which cannot be zero?A. 10,10,10B. 10,10,20C. 10,20,20D. 10,20,40 |
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Answer» Correct Answer - D For the resultant of some vectors to be zero, they should from a closed figure taken in the same order. |
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| 596. |
Two vectors `vecP and vecQ` lie one plane. Vectors `vecR` lies in a differenct plane. In such a case, `vecP + vecQ + vecR`A. can be zeroB. cannot be zeroC. lies in the sama plane as `vecP` or `vecQ`D. lies in the plane different from that of any two of 3 vectors |
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Answer» Correct Answer - B::C Vector properties. |
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| 597. |
We can order events in time and there is a sense of time, distinguishing past,present and futher. Is therefore, time a vector? |
| Answer» Time always flows from past to present and then to future, so a direction can be assigned to time. However, as the direction is unique, it is not to be stated ,i.e.specified. As the direction is not to be specified, time cannot be a vector though it has a direction. | |
| 598. |
Can a vector be zero if any of its components is not zero? |
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Answer» In terms of components in three dimestions, `vecA = hati A_(x) + hatj A_(y) + hatk A_(z)` with `A = sqrt(A_(x)^(2)+ A_(y)^(2) + A_(z)^(2))` so if any of `A_(x),A_(y) and A_(z)` is not zero (or null) all its components must vanish. |
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| 599. |
If `|vecA xx vecB| = sqrt(3) vecA .vecB` then the value of `|vecA xx vecB|` is : |
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Answer» Correct Answer - B `ABsin theta= Ab cos thetasqrt(3)` `tan theta = sqrt(3) " " rArr theta = 60^(@)` `|(vecA + vecB)|= sqrt(A^(2)+ B^(2)+ 2AB cos 60^(@))` |
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| 600. |
The angle between `vecA = hati + hatj and vecB = hati - hatj` isA. `45^(@)`B. `90^(@)`C. `-45^(@)`D. `180^(@)` |
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Answer» Correct Answer - B (b ) Given ,`A=hati +hatj` and `B=hati-hatj` We know that A.B=|A||b|cos theta ` implies (hati+hatj).(hati-hatj)=(sqrt(1+1))(sqrt(1+1)) costheta` where `theta` is the angle between A and B `implies cos theta=(1-0+0-1)/(sqrt(2)sqrt(2))=(0)/(2)=0` `implies theta90^(@)` |
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