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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.
| 601. |
If 2` vec A C`= 3` vec C B`, then prove that 2` vec O A`=3` vec C B`then prove that 2` vec O A`+ 3` vec O B`=5` vec O C`where `O`is the origin. |
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Answer» `2vec(AC)=3vec(CB)` or `2(vec(OC)-vec(OA))=3(vec(OB)-vec(OC))` or `" "2vec(OA)+3vec(OB)=5vec(OC)` |
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| 602. |
Prove that points `hati+2hatj-3hatk, 2hati-hatj+hatk and 2hati+5hatj-hatk` form a triangle in space. |
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Answer» Given points are `A(hati+2hatj-3hatk), B(2hati-hatj+hatk), C(2hati+5hatj-hatk)` `" ""Vectors "vec(AB)=hati-3hatj+4hatk and vec(AC)=hati+3hatj+2hatkk` Clearly vectors `vec(AB) and vec(AC)` are non-collinear as there does not exist any real `lamda` for which `vec(AB)=lamdavec(AC)`. `" "` Hence, vecors `vec(AB) and vec(AC)` or the given three points form a triangle. |
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| 603. |
Two equal vector have a resultant equal to either of them, then the angle between them will be:A. `60^(@)`B. `120^(@)`0C. `90^(@)`D. `0^(@)` |
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Answer» Correct Answer - B (b) We have R=A=B `thereforeR^(2)=R^(2)+R^(2)+2R Rcos theta ` `thereforecos theta=-(1)/(2) ` `thereforetheta=120^(@)` |
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| 604. |
What is the angle between `hati+hatj+hatk and hatj`?A. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. none of these |
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Answer» Correct Answer - D (d ) Angle `theta cos^(-1) ((A.B)/(AB))=cos^(-1) ((1)/(sqrt(3)))` |
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| 605. |
If a vector `2hat(i)+3hat(j)+8hat(k)` is perpendicular to the vector `4hat(j)-4hat(i)+alpha hat(k)`. Then the value of `alpha` isA. -1B. `1/2`C. `-1/2`D. 1 |
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Answer» Correct Answer - C We are given `vec(a)= 2hat(i)+3hat(j)+8hat(k), vec(b)= 4hat(j)-4hat(i)+alpha hat(k)` According to the above hypothesis, If `vec(a)_|_ vec(b)`, then dot product of Vector should be zero `vec(a).vec(b)=0` `implies (2hat(i)+3hat(j)+8hat(k))(-4hat(i)+4hat(j)+alpha hat(k))=0` `implies -8+12+8 alpha=0implies 8 alpha = -4` `:. alpha= -4/8= -1/2` |
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| 606. |
If `vec(A)=4hat(i)-3hat(j)` and `vec(B)=6hat(i)+8hat(j)`,then find the magnitude and direction of `vec(A)+vec(B)`. |
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Answer» `vec(A)+vec(B)=4hat(i)-3hat(j)+6hat(i)+8hat(j)=10hat(i)+5hat(j)` `|vec(A)+vec(B)|=sqrt((10^(2))+(5^(2)))=5sqrt(5)` `tan theta=5/(10)=1/2 rArrtheta=tan^(-1)(1/2)` |
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| 607. |
Unit vector parallel to the resultant of vectors `vec(A)= 4hat(i)-3hat(j)` and `vec(B)= 8hat(i)+8hat(j)` will beA. `(24hat(i)+5hat(j))/(13)`B. `(12hat(i)+5hat(j))/(13)`C. `(6hat(i)+5hat(j))/(13)`D. None of these |
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Answer» Correct Answer - B Resutant of vectors `vec(A)` and `vec(B)` `vec(R )= vec(A)+vec(B)= 4hat(i)-3hat(j)+8hat(i)+8hat(j)= 12hat(i)+5hat(j)` `hat(R )= (vec(R ))/(|R|)= (12hat(i)+5hat(j))/(sqrt((12)^(2)+(5)^(2)))= (12hat(i)+5hat(j))/(13)` |
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| 608. |
The angle between `(vecAxxvecB)` and `(vecBxxvecA)` is (in radian)A. `pi//2`B. `pi`C. `pi//4`D. zero |
| Answer» Correct Answer - B | |
| 609. |
If `vecA` and `vecB` are two vectors, then which of the following is wrong?A. `vecA.vecB=vecB.vecA`B. `vecA+vecB=vecB+vecA`C. `vecAxxvecB=vecBxxvecA`D. `vecAxxvecB=-vecBxxvecA` |
| Answer» Correct Answer - C | |
| 610. |
Two vectors `vecA` and `vecB` are such that `vecA+vecB=vecA-vecB`. ThenA. `vecA.vecB=0`B. `vecAxxvecB=0`C. `vecA=0`D. `vecB=0` |
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Answer» Correct Answer - D |
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| 611. |
If a vector `2hati+3hatj+8hatjk` is perpendicular to the vector `4hatj-4hati+alphahatk`. Then the value of `alpha` isA. `-1`B. `1/2`C. `-1/2`D. `1` |
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Answer» Correct Answer - C |
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| 612. |
If a vector `2hati +3hatj +8hatk` is perpendicular to the vector `4hati -4hatj + alphahatk,` then the value of `alpha` isA. `1//2`B. `-1`C. `-1//2`D. 1 |
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Answer» Correct Answer - C (c ) Dot of these two vector should be equal to zero ad sthety are perpendiculaar to each other . `therefore A.B=-8+12+8alpha=0` `8alpha=-4=. alpha =(-1)/(2)` |
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| 613. |
The unit vectors perpendicular to `vecA=2hati+3hatj+hatk` and `vecB=hati-hatj+hatk` isA. `(4hati-hatj-5hatk)/sqrt(42)`B. `(4hati-hatj+5hatk)/sqrt(42)`C. `(4hati+hatj+5hatk)/sqrt(42)`D. `(4hati+hatj-5hatk)/sqrt(42)` |
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Answer» Correct Answer - A `hatn=(vecAxxvecB)/(|vecAxxvecB|)~ |
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| 614. |
Dot product is used in the determination of (a) Work done by a force (b) Power developed by an automobile moving with uniform velocity. (c) The normal flux linked with a coil kept in magnetic field. (d) The force acting on a conductor carrying current kept in a magnetic fieldA. a,d are trueB. b,d are trueC. a,b,c are trueD. c,d are true |
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Answer» Correct Answer - C (a) `W=vecF.vecS,` (b) `P=vecF.vecv` (c) `phi=vecB.vecA,` (b) `vecF=idveclxxvecB` |
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| 615. |
If `vecA=4hati-3hatj` and `vecB=6hati+8hatj` then magnitude and direction of `vecA+vecB` will beA. `5, tan^(-1)(3//4)`B. `5sqrt(5),tan^(-1)(1//2)`C. `10,tan^(-1)(5)`D. `25, tan^(-1)(3//4)` |
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Answer» Correct Answer - B |
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| 616. |
If `vecA=4hati-3hatj` and `vecB=6hati+8hatj` then magnitude and direction of `vecA+vecB` will beA. `5, tan^(-1)(3//4)`B. `5sqrt(5), tan ^(-1)(1//2)`C. `10, tan ^(-1)(5)`D. `25 ,tan ^(-1)(3//4)` |
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Answer» Correct Answer - B (b) `A+B=10hati+5hatj` ` therefore |A+B|=sqrt(100+25)=5sqrt(5)` Angle of A.B with X-axis, `theta=tan^(-1)((5)/(10))=tan^(-1)((1)/(2))` |
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| 617. |
If `vecA=4hati-3hatj` and `vecB=6hati+8hatj` then magnitude and direction of `vecA+vecB` will beA. `5,tan^(-1)(3//4)`B. `5sqrt(5) ,tan^(-1)(1//2)`C. `10, tan^(-1)(5)`D. `25,tan^(-1)(3//4)` |
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Answer» Correct Answer - B (b)` A+B =10hati +5hatj` `therefore|A+B|=sqrt((10^(2)+(5)^(2))=5sqrt(5)` Angle of (A+B) with Xasis ,`theta=tan^(-1)((5)/(10))=tan^(-1)((1)/(2))` |
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| 618. |
Which of the following is not true? If `vecA=3hati+rhatj` and `vecB=6hati+8hatj` where `A` and `B` are the magnitude of `vecA` and `vecB`A. `vecAxxvecB=0`B. `A/B=1/2`C. `vecA.vecB=50 A=5`D. none |
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Answer» Correct Answer - D |
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| 619. |
A 120 m long train is moving towards west with a speed of 10 `m//s`. A bird flying towards east with a speed of 5 `m//s` crosses the train. The time taken by the bird to cross the train will beA. 16 secB. 12 secC. 10 secD. 8 sec |
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Answer» Correct Answer - D |
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| 620. |
The car makes a displacement of 100m towards east and then 200m towards north. Find the magnitude and direction of the resultant.A. 223,7m, `tan^(-1)(2),N` of EB. 223,7m, `tan^(-1)(2),E` of NC. 300m, `tan^(-1)(2)`, N of ED. 100m, `tan^(-1)(2)`, N of E |
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Answer» Correct Answer - A `s_(1)=100hati,s_(2)=200hatj` `s=sqrt(s_(1)^(2)+s_(2)^(2)), Tan theta=(s_(2))/(s_(1))rarrN` of E |
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| 621. |
The component of vector `A=2hati+3hatj` along the vector `hati+hatj` isA. `(5)/(sqrt(2))`B. `4sqrt(2)`C. `(sqrt(2))/(3)`D. None of these |
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Answer» Correct Answer - A (a) component of A along B `=acostheta(A.B)/(B)=(2+3)/(sqrt(1+1))=(5)/(sqrt(2))` |
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| 622. |
Find the components of a vector `A = 2hati + 3hatj` along the directions of `hati + hatj and hati - hatj.` |
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Answer» `vecC=(A cos theta)hatB=(vecA.vecB)/(|vecB|)[(vecB/(|vecB|)]` `vecC=[(2+3)/(sqrt(1+1))][(hati+hatj)/(sqrt(1+1))]=(5)/(2)(hati+hatj)`. |
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| 623. |
The torque of the force `vecF=(2hati-3hatj+4hatk)N` acting at the point `vecr=(3hati+2hatj+3hatk)m` about the origin beA. `6hati - 6 hatj +12 hatk`B. `17hati - 6 hatj -13 hatk`C. `-6hati + 6 hatj -12 hatk`D. `-17hati + 6 hatj -13 hatk` |
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Answer» Correct Answer - B `vectau = vecr xx vecF` `=|{:(hati,hatj,hatk),(3,2,3),(2,-3,4):}|` `= hati ( 8+ 9) - hatj (12 - 6) + hatk(-9-4)` `= 17 hati - 6 hatj - 13 hatk` |
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| 624. |
If `vecF=hati+2hatj-3hatk` and `vecr=2hati-hatj+hatk` find `vecrxxvecF` |
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Answer» `vecrxxvecF=|:(i,j,k),(2,-1,1),(1,2,-3):|` `(3-2)hati-(-6-1)hati+(4+1)hatk=hati+7hatj+shatk` |
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| 625. |
Let `veca=hati + hatj +hatk,vecb=hati- hatj + hatk and vecc= hati-hatj - hatk` be three vectors. A vectors `vecv` in the plane of `veca and vecb` , whose projection on `vecc is 1/sqrt3` is given byA. `hati-3hatj+3hatk`B. `-3hati-3hatj-hatk`C. `3hati-hatj+3hatk`D. `hati+3hatj-3hatk` |
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Answer» Correct Answer - C Let `V=a+ lamda b` `v=(1+lamda )hati +(1-lamda ) hatj +(1+lamda )hatk` projection of v on c`=(v.c)/(|c|)=(1)/(sqrt(3))` `implies ((1+lamda )-(1-lamda )-(1+lamda ))/(sqrt(3))=(1)/(sqrt(3))` `implies lamda =2` `therefore v= 3 hati - hatj + 3 hatk` |
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| 626. |
If a particle moves from the point `A(1 , 2 , 3)` to the point `B( 4 , 6 ,9)`, its displacement vector be |
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Answer» The particle is moving from point A(1,2,3) to the point B(4,6,9) ` therefore`According to the formula of displacement vector `D=(x_(2-X_(1))hati+ (y_(2)-y_(1)) hatj +(z_(2)-z_(1)) hatk` where ,`x_(1)y_(1,z_(1=1,2,3` ` x_(2),y_(2),Z_(3)=4,6,9` putting these values , in Eq .(i) We get ` d=(4-1)hati+(6-2)hatj+(9-3)hatk` `=3hati +4hatj +6hatk` |
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| 627. |
If a particle moves from the point `A(1 , 2 , 3)` to the point `B( 4 , 6 ,9)`, its displacement vector beA. `hat(i) + hat(j) + hat(k)`B. `2 hat(i) + 3 hat(j) + 4hat(k)`C. `3hat(i) + 4hat(j) + 6hat(k)`D. `3hat(i) + 4hat(j) + 5hat(k)` |
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Answer» Correct Answer - C `vec(r_(1)) = I + 2j + 3k , vec_(r_(2)) = 4i + 6j + 9k` `vec(d) = vec(r_(2)) - vec(r_(1)) = 3 I + 4j + 6k` |
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| 628. |
If a particle moves from point `P(2,3,5)` to point `Q(3,4,5)`. Its displacement vector beA. `hat(i)+hat(j)+10hat(k)`B. `hat(i)+hat(j)+5hat(k)`C. `hat(i)+hat(j)`D. `2hat(i)+4hat(j)+6hat(k)` |
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Answer» Correct Answer - C Displacement vector `vec(r )= Deltaxhat(i)+Deltayhat(j)+Deltazhat(k)` `=(3-2)hat(i)+(4-3)hat(j)+(5-5)hat(k)= hat(i)+hat(j)` |
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| 629. |
Find the moment of force `vecF = hati + hatj + hatk` acting at point `(-2, 3, 4)` about the point `(1, 2, 3)`. |
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Answer» Correct Answer - `4(hati -hatk)` `vecF = hati + hatj +hatk` Let `vecr_(2) = (-2hati + 3hatj + 4hatk)` `" "vecr_(1) =(hati + 2hatj + 3hatk)` `vecr= vecr_(2) - vecr_(1) = -3hati +hatj +hatk` Moment of force `= vectau = vecr xx vecF` `" "=(1 -1)hati - (-3-1) hatj + (-3-1)hatk` `" "= - 4hatj + 4hatk` |
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| 630. |
Compute the force (in vetcor notation ) on an electron moving with velocity `vecV = 2.5 xx 10^(6)hati m//s` in a magnetic field `vecB =(10hati- 6 hatk) xx 10 ^(2) Wb//m^(2)`, if charge on an electron `e =1.6 xx 10^(-19)` coulomb. |
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Answer» Correct Answer - `vecF= 2.4 xx 10^(-10)hatj` newton `vecF= q (vecv xx vecB)` `=q[(2.5 xx 10^(6))hati xx (10 xx 10^(2)) hati- (6 xx 10^(2))hatk]N` `= 1.6xx 10 ^(-19) xx 15xx 10^(8) hatj N` `= 2.4 xx 10^(-10)hatjN` |
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| 631. |
If a particle moves from point `P(2,3,5)` to point `Q(3,4,5)`. Its displacement vector beA. `hati+hatj+10hatk`B. `hati+hatj+5hatk`C. `hati+hatj`D. `2hati+4hatj+6hatk` |
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Answer» Correct Answer - C (c ) Displacement vector `S=r_(Q)-r_(p)` `=(3hati+4hatj+5hatk)-(2hati+3hatj+5hatk)=hati+hatj` |
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| 632. |
If two waves of same frequency and same amplitude superimpose and produce third wave of same amplitude, then waves differ in phase by –A. zeroB. `pi`C. `pi//2`D. `pi//4` |
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Answer» Correct Answer - D `A_(R)=sqrt(A_(1)^(2) + A_(2)^(2) + 2A_(1)A_(2) cos theta)` where `theta ` is the phase difference between the waves but `A_(R) = A_(1) = A_(2) = A` So, `" "A^(2) = A^(2) + 2A^(2) cos theta` `" "cos theta = -(1)/(2)` `" " theta = -(2pi)/(3)` |
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| 633. |
A force `2hati+hatj-hatk` newton acts on a body which is initially at rest.If the velocity of the body at the end of `20seconds` is `4hati+2hatj+2hatk ms^(-1)`, the mass of the bodyA. 20kgB. 15kgC. 10kgD. 5kg |
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Answer» Correct Answer - C `vecFt=mvecV` |
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| 634. |
Two trains along the same straight rails moving the constant speed `60km//hr` and `30km//hr` respectively towards each other. If at time `t=0`, the distance between then is `90km`, the time when they collide isA. 1 hrB. 2 hrC. 3 hrD. 4 hr |
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Answer» Correct Answer - A |
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| 635. |
A steam boat goes across a lake and comes back : (a) on a quiet day when the water is still and (b) on a rough day when there is a uniform current so as to help the journey onward and to impede the journey backward. If the speed of launch on both days same, in which case will it complete the journey in lesser time?A. Case (a)B. Case (b)C. Same in bothD. Nothing can be predicted |
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Answer» Correct Answer - A |
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| 636. |
Two equal forces (P each) act at a point inclined to each other at an angle of `120^@`. The magnitude of their resultant isA. `P//2`B. `P//4`C. `P`D. `2P` |
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Answer» Correct Answer - C |
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| 637. |
Let the angle between two nonzero vector `vecA and vecB` is `120^@` and its resultant be `vecC`.A. `vecC` must be equal to `|vecA-vecB|`B. `vecC` must be less than `|vecA-vecB|`C. `vecC` must be greater than `|vecA-vecB|`D. `vecC` may be equal to `|vecA-vecB|` |
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Answer» Correct Answer - C |
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| 638. |
Assertion: Minimum number of non-equal Vectors in a plane required to give zero resultant is three. Reason: If `vec(A)+vec(B)+vec(C )= vec(0)`, then they must lie in one planeA. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false. |
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Answer» Correct Answer - B |
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| 639. |
If the angle between two vectors A and B is `120^(@)`, then its resultant C will beA. `C=|A-B|`B. `Clt|A-B|`C. `Cgt|A-B|`D. `C=|A+B|` |
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Answer» Correct Answer - C (c ) Resultant of two vector lies between `|A+B|and |A-B|` |
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| 640. |
The component of a vector isA. Always less than its magnitudeB. Always greater than its magnitudeC. Always equal to its magnitudeD. None of these |
| Answer» Correct Answer - D | |
| 641. |
Find the unit vector parallel to the resultant of the unit vectors `vec(A) = 2 hati - 6 hatj - 3hatk` and `vec(B) = 4 hati +3hatj - hatk`. |
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Answer» Correct Answer - `1//sqrt(61) (6hati - 3hatj - 4hatk)` |
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| 642. |
Let the angle between two non - zero vectors `vec(A)` and `vec(B)` be `120^(@)`and its resultant be `vec( C)`.A. `C` must be equal to `|vec(A) - vec(B)|`B. `C` must be less than `|vec(A) - vec(B)|`C. `C` must be greater than `|vec(A) - vec(B)|`D. `C` may be equal to `|vec(A) - vec(B)|` |
| Answer» Correct Answer - B | |
| 643. |
What is the angle between `vec(P)` and the resultant of `(vec(P)+vec(Q))` and `(vec(P)-vec(Q))` ?A. zeroB. `tan^(-1) (p//Q)`C. `tan ^(-1)(Q//p)`D. `tan^(-1)(P-Q)//(P+Q)` |
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Answer» Correct Answer - A (a) Resultant of `(P+Q)and (P-Q)is P+Q+P-Qor 2P` which is parallel to p. So angles between P and 2P will be zero. |
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| 644. |
If `F_(1) and F_(2)` aare two vectors of equal magniudes F such that `|F_(1).F_(2)|= |F_(1)xxF_(2)|, "then"|F_(1) +F_(2)|` equals toA. `sqrt((2+sqrt(2)))F`B. `2F`C. `Fsqrt(2)`D. None of these |
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Answer» Correct Answer - A (a) `F.F.costheta =F.F sin theta `or tantheta `=1or theta=45^(@)` `|F_(1).F_(2)|=sqrt(F^(2)+F^(2)+2,FFcos 45^(@))` `(sqrt(2+sqrt(2)))F` |
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| 645. |
The positoin vector of a particle is `vecr=(acos omegat)hati+(asinomegat)hatj`. The velocity of the paritcle isA. Parallel to the position vectorB. Perpendicular to the position vectorC. Directed towards the originD. Directed away from the origin |
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Answer» Correct Answer - B |
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| 646. |
The area of the parallelogram whose diagonals are the vectors `hati+3hatj- 2hatk and 3hati+hatj-4hatk`A. `3sqrt(42)` sq unitsB. `sqrt(42)` sq untisC. `2sqrt(42)` sq unitsD. None of these |
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Answer» Correct Answer - A Let `a=hat(i)+3hat(j)-2hat(k) and b=3hat(i)+hat(j)-4hat(k)` The `a xxb=|{:(hat(i),hat(j),hat(k)),(1,3,-2),(3,1,-4):}|` `=hat(i)(-12+2)-hat(j)(-4+6)+hat(k)(1-9)` `=-10hat(i)-2hat(j)-8hat(k)` `therefore` Area of parallelogram `=(1)/(2)|axxb|` `=(1)/(2)sqrt((-10)^2+(-2)^2+(-8)^2)` `=(1)/(2)sqrt(100+4+64)=(1)/(2)sqrt(168)` `=sqrt(42)` sq.units. |
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| 647. |
The area of the parallelogram whose adjacent sides are `hati-3hatj+hatk and hati+hatj+hatk` isA. `4sqrt2`B. `-4sqrt2`C. `4sqrt3`D. None of these |
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Answer» Correct Answer - B Let , `a=hat(i)-3hat(j)+hat(k) and b=hat(i)+hat(j)+hat(k)` `therefore` Area of parallelogram having adjacent sides, a and b is `|a xxb|=|{:(hat(i),hat(j),hat(k)),(1,-3,1),(1,1,1):}|` `=|hat(i)(-3-1)-hat(j)(1-1)+hat(k)(1+3)|` `=|-4hat(i)+4hat(k)=sqrt((-4)^2+4^2)=sqrt(16+16)` `=sqrt(32)=4sqrt(2)` sq units. |
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| 648. |
If `a=hati+2hatj+3hatk` and `b=hatixx(axxhati)+hatjxx(axx hatj)+hatk+(a xx hatk)`, then length of b is equal toA. `sqrt(12)`B. `2sqrt(12)`C. `3sqrt(14)`D. `2sqrt(14)` |
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Answer» Correct Answer - B We have , `a=hat(i)+2hat(j)+3hat(k)` `b=hat(i)xx(axxhat(i))+hat(j)xx(axxhat(j))+hat(k)xx(axxhat(k))` Now , `hat(i)xx(axxhat(i))=(hat(i).hat(i))a-(hat(i).a)hat(i)` `=a-a_(1)hat(i)" "["let" (a=a_(1)hat(i)+a_(2)hat(j)+a_(3)hat(k))]` Similarly , `hat(j)xx(axxhat(i))=a-a_(2)hat(j) and hat(k)xx(axxhat(k))` `therefore b=3a-a=2a=2(hat(i)+2hat(j)+3hat(j))` `rArr |b|=sqrt(4+16+36)=sqrt(56)=2sqrt(14)` . |
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| 649. |
If the vectors of a trianlge are `A(hati+hatj+2hatk), B(3hati-hatj+2hati) and C(2hati-hatj+hatk)` the area of triangle isA. `2sqrt3` sq unitsB. `sqrt3` sq unitsC. `2sqrt3` sq unitsD. 3 sq units |
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Answer» Correct Answer - D Now , `AB=(3hat(i)-hat(j)+2hat(k))-(hat(i)+hat(j)+2hat(k))=2hat(i)-2hat(j)` and `AC=(2hat(i)-hat(j)+hat(k))-(hat(i)+hat(j)+2hat(k))=hat(i)-2hat(j)-hat(k)` `(ABxxAC)=|{:(hat(i),hat(j),hat(k)),(2,-2,0),(1,-2,-1):}|` `=hat(i)(2-0)-hat(j)(-2-0)+hat(k)(-4+2)` `=2hat(i)+2hat(j)+2hat(k)` `therefore` Area of triangle `AB=(1)/(2)|ABxxAC|` `=(1)/(2)sqrt((2)^2+(2)^2+(-2)^2)` `=(2)/(2)sqrt(1+1+1)=sqrt(3)` sq units. |
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| 650. |
The vectors `a=2hati+hatj-2hatk, b=hati+hatj`. If c is a vector such that `a.c=|c| and |c-a|=2sqrt2,` angle between `axxb` and c is `45^(@)`, then `|(axxb)xxc|` isA. 3B. `(sqrt3)/(2)`C. `(3sqrt2)/(2)`D. None of these |
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Answer» Correct Answer - C Now, `|a|^(2)=9 and |b|^(2) = 2 ` `therefore |c-a|^(2) =|c|^(2) +|a|^(2) -2c*a=8` `=|c|^(2)+9-2|c| =8 rArr |c| = 1` Now , ` a xx b = | (hati , hatj , hatk ) , ( 2, 1 , -2), (1, 1 , 0)|=2 hati - 2 hatj+ hatk ` ` rArr | a xx b| = sqrt( 2^(2) +(-2)^(2)+1^(2))=3` `therefore |(a xx b) xx c|=|a xx b||c|sin 45^(@) = 3(1) ((1)/(sqrt(2)))=(3sqrt(2))/(2)` |
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