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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.
| 151. |
If a and b are unit vectors, then what is the angle between a and b for `sqrt3a-b` to be a unit vector?A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - A We have, `(sqrt(3)a-b)^(2)+3a^(2)+b^(2)-2sqrt(3)a.b` `implies a.b =sqrt(3)/2 implies cos theta=sqrt(3)/2 implies theta = 30^(@)` |
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| 152. |
In a righht angled triangle the three vectors `veca,vecb` and `vecc` add to zero. Then `veca,vecb` is A. `-9`B. `+9`C. 0D. `-3` |
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Answer» Correct Answer - C `veca.vecb=ab cos 90^(@)` |
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| 153. |
A vector parallel to the vector `(hati+2hatj)` and having magnitude `3sqrt(5)` units isA. `3hati+6hatj`B. `6hati-3hatj`C. `4hati-2hatj`D. `hati-2hatj` |
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Answer» Correct Answer - B `veca.vecb=0,|vecb|=3sqrt(5)` |
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| 154. |
If the scalar product of the vector `hati+hatj+2hatk` with the unti vector along `mhati+2hatj+3hatk` is equal to 2, then one of the value of m isA. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - D `(hati +hatj +2 hatk ). ((mhati +2hatj +3hatk )/( sqrt(m^(2)+2^(2)+3^(2))))=2` `implies m+2+6=2sqrt(13+m^(2))` `implies (m+8)^(2)=4(13+m^(2))` `implies m^(2)+16m+64 =4m^(2)+52` ` implies 3m^(2)-16m - 12=0 ` `implies (3m+2)(m-6)=0` `implies m=6,-(2)/(3)` |
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| 155. |
If `vec(A) = 2hati +3 hatj +hatk` and `vec(B) = 3hati + 2hatj + 4hatk`, then find the value of `(vec(A) +vec(B)) xx (vec(A) - vec(B))` |
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Answer» Correct Answer - `-20 hati +10 hatj +10 hatk` |
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| 156. |
Find `(2A)xx(-3B)`, if `A =2hati - hatj` and `B =(hatj+hatk)` |
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Answer» Correct Answer - A::B `2A = 4hati - 2hatj` `-3B =- 3hatj - 3hatk` `rArr (2A)xx(-3B) = |(hati, hatj ,hatk),(4,-2,0),(0,-3,-3)|` `=hati(6-0) +hatj (0+12)+hatk(-12 - 0 ) = 6hati+12hatj - 12hatk` . |
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| 157. |
If the angle between the vectors A and B is `theta`, the value of the product(BxA)). A is equal toA. `BA^2 cos theta`B. `BA^2 sin theta`C. `BA^2 sin theta cos theta`D. zero |
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Answer» Correct Answer - D `(BxxA) is _|_` to both A and B `(BxxA) .A = 0.` |
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| 158. |
Given that P 12, Q = 5 and R = 13 also P+Q=R, then the angle between P and Q will beA. `pi`B. `(pi)/(2)`C. zeroD. `(pi)/(4)` |
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Answer» Correct Answer - B `R = sqrt(P^2 + Q^2 + 2PQ cos theta)` Substituting the values of P,Q and R we get, `cos theta = 0` `rArr theta = 90^@.` |
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| 159. |
The angles between P+Q and P-Q will beA. `90^@`B. between `0^@ and 180^@`C. `180^@` onlyD. None of these |
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Answer» Correct Answer - B (P+Q).(P-Q) `=P^2 + PQ cos theta - PQ cos theta -Q^2` `=P^2 - Q^2` Since dot product may be positive (if P> Q), negative (if Q>P) or zero (if P=Q). Therefore angle between (P+Q) and (P-Q) may be acute, obtuse or `90%.` |
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| 160. |
The angles between the two vectors `vecA=3hati+4hatj+5hatk` and `vecB=3hati+4hatj-5hatk` will beA. `60^(@)`B. ZeroC. `90^(@)`D. None of these |
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Answer» Correct Answer - D |
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| 161. |
Given `p=3hat(i)+2hat(j)+4hat(k), a=hat(i)+hat(j), b=hat(j)+hat(k), c=hat(i)+hat(k)` and `p=x a +y b +z c`, then x, y and z are respectivelyA. `3/2, 1/2, 5/2`B. `1/2, 3/2, 5/2`C. `5/2, 3/2, 1/2`D. `1/2, 5/2, 3/2` |
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Answer» Correct Answer - B `p=x a + yb +zc` `rArr" "3hati+2hatj+4hatk=x(hati+hatj)+y(hatj+hatk)+z(hati+hatk)` `rArr" "3hati+2hatj+4hatk=(x+z)hati+(x+y)hatj+(y+z)hatk` `x+z=3" ...(i)"` `x+y=2" ...(ii)"` `x+z=4" ...(iii)"` Now, subtracting Eq. (ii) from Eq. (i), we get `z-y=1" ...(iv)"` From Eqs. (iii) and (iv), we gt `z=5/2, y=3/2` From Eq. (ii) we get `x=1/2` Hence, `x=1/2, y=3/2, z=5/2` |
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| 162. |
A particle has initial velocity `(2hati+3hatj)` and acceleration `(0.3hati+0.2hatj)`. The magnitude of velocity after 10 second will beA. 9 unitsB. `9sqrt(2) units`C. `5sqrt(2)units`D. 5 unit |
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Answer» Correct Answer - C `vec(v)= vec(u)+vec(a)t` `=(2hat(i)+3hat(j))+(0.3hat(i)+0.2hat(j))xx10` `=(5hat(i)+5hat(j))implies |vec(v)|=5sqrt(2)units` |
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| 163. |
If a, b and c are the three vectors mutually perpendicular to each other to form a right handed system and `|a|=1, |b|=3 and |c|=5`, then `[a-2b b-3c c-4a]` is equal to |
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Answer» Correct Answer - C Given that, `|a|=1, |b|=3` and `|c|=5` `:. [(a-2b,b-3c,c-4a)]` `=(a-2b).{(b-3c)xx(c-4a)}` `=(a-2b). {bxxc-4bxx +12 xx}` `=(a-2b).(a+4c+12b)` `=a.a-24b.b=1-24xx9` `=1-216=-215` |
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| 164. |
Find the sine of the angle between the vectors `veca=3hati+hatj+2hatkandvecb=2hati-2hatj+4hatk`A. `sqrt((2)/(7))`B. `(2)/(sqrt7)`C. `(sqrt2)/(7)`D. None of these |
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Answer» Correct Answer - B Given that, `a=3 hat(i)+hat(j)+2hat(k)` and `b=2 hat(i)-2hat(j)+4hat(k)` Let `theta` be the angle between them. `:. cos theta=(a.b)/(|a||b|)=((3hat(i)+hat(j)+2hat(k)))/sqrt(9+1+4)((2hat(i)-2hat(j)+4hat(k)))/sqrt(4+4+16)` `implies cos theta=(6+(-2)+8)/(sqrt(14)sqrt(24))=12/(4sqrt(21))` `:. sin theta=sqrt(1-cos^(2) theta)` `=sqrt(1-144/336)=sqrt(192/336)` `implies sin theta =2/sqrt(7)` |
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| 165. |
If `|veca|=3` and `-1 |
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Answer» Correct Answer - B The smalest value of |k | will exist at numerically smallest value of k, i.e, at k =0, which gives `|ka|=|k||a|=0xx3=0` The numerically greaterst value of k is 2 at which `|k a|=6` |
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| 166. |
If ABCDEF is a regular hexagon then `vec(AD)+vec(EB)+vec(FC)` equals : |
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Answer» Correct Answer - D ABCDEF is a regular hexogen. We know from the hexagon that AD is parallel to BC `implies AD=2BC` Similarly, EB is parallel to FA `implies EB = 2FA` and FC is parallel to AB `impliesFC =2AB` Thus, `AD+EB +FC =2BC+2FA+2AB` `=2(FA+AB+BC)` `=2(FC=2(2AB)=4AB)` |
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| 167. |
If ` veca and vecb` are unit vectors inclined to x-axis at angle ` 30^(@) and 120^(@)` then ` |veca +vecb|` equalsA. `sqrt((2)/(3))`B. `sqrt2`C. `sqrt3`D. 2 |
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Answer» Correct Answer - C Clearly, angle between a and `b=pi/2` `implies a.b=0` `:. |a+b|^(2)=a^(2)+b^(2)+2a.b=1+1+0=2` `implies |a+b|=sqrt(2)` |
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| 168. |
In figure, `vec(E )` equals A. `vec(A)` and `vec(B)` are perpendicular to each otherB. `vec(B)`C. `vec(A)+vec(B)`D. `-(vec(A)+vec(B))` |
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Answer» Correct Answer - D From right triangle `vec(A)+vec(B)+vec(E )=0` `implies vec(E )= -(vec(A)+vec(B))` |
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| 169. |
Let `vec(C )= vec(A)+vec(B)` thenA. `|vec(C )|` is always greater then `|vec(A)|`B. It is possible to have `|vec(C )|lt|vec(A)| and |vec(C )lt|vec(B)|`C. `C` is always equal to `A+B`D. `C` is never equal to `A+B` |
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Answer» Correct Answer - B `vec(C )+vec(A)= vec(B)` The value of `C` lies between `A-B` and `A+B` `:. |vec(C )|lt|vec(A)| or |vec(C )|lt|vec(B)|` |
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| 170. |
A body is at rest under the action of three forces, two of which are `vec(F)_(1)= 4hat(i), vec(F)_(2)= 6hat(j)`, the third force isA. `4hat(i)+6hat(j)`B. `4hat(i)-6hat(j)`C. `-4hat(i)+6hat(j)`D. `-4hat(i)-6hat(j)` |
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Answer» Correct Answer - D `vec(F)_(1)+vec(F)_(2)+vec(F)_(3)=0 implies 4hat(i)+6hat(j)+vec(F)_(3)=0` `:. Vec(F)_(3)= -4hat(i)-6hat(j)` |
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| 171. |
Two forces `F_(1)=1N` and `F_(2)=2N` act along the lines x=0 and y=0, respectively. Then find the resultant of forces.A. `hat(i)+2hat(j)`B. `hat(i)+hat(j)`C. `3hat(i)+2hat(j)`D. `2hat(i)+hat(j)` |
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Answer» Correct Answer - D `x=0` means y-axis implies `vec(F)_(1)= hat(j)` y=0 means x-axis `implies vec(F)_(2)= 2hat(i)` So resultant `vec(F)= vec(F)_(1)+vec(F)_(2)= 2hat(i)+hat(j)` |
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| 172. |
Following forces start acting on a particle at rest at the origin of the co-ordinate system simultaneously `vec(F)_(1)= -4hat(i)-5hat(j)+5hat(k), vec(F)_(2)= 5hat(i)+8hat(j)+6hat(k), vec(F)_(3)= -3hat(i)+4hat(j)-7hat(k)` and `vec(F)_(4)= 2hat(i)-3hat(k)` then the particle will moveA. in x-y planeB. In y-z planeC. In x-z planeD. Along x-axis |
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Answer» Correct Answer - B `vec(F)_(1)+vec(F)_(2)+vec(F)_(3)+vec(F)_(4)` `=(-4hat(i)+5hat(j)-3hat(i)+2hat(i))+(-5hat(j)+8hat(j)+4hat(j)-3hat(j))` `+(5hat(k)+6hat(k)-7hat(k)-2hat(k))=4hat(j)+2hat(k)` `:. The particle will move in y-z plane. |
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| 173. |
A swimmer can swim in still water with speed `v` and the river is flowing with velociyt `v//2`. To cross the river in shortest distance, he should swim making angle `theta` with the upstream. What is the ratio of the time taken to swim across the shortest time to that is swimming across over shortest distanceA. `costheta`B. `sin theta`C. `tan theta`D. `cot theta` |
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Answer» Correct Answer - A |
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| 174. |
The horizontal component of the weight of a body of mass m isA. mgB. `(mg)/(2)`C. zeroD. inifinity |
| Answer» Correct Answer - C | |
| 175. |
Three vectors satisfy the relation A.B =0 and A.C=0 then A is parallel toA. `vecC`B. `vecB`C. `vecBxxvecC`D. `vecB`.`vecC` |
| Answer» Correct Answer - C | |
| 176. |
Cross product of vectors obeysA. commutative lawB. associative lawC. distributive lawD. all the above |
| Answer» Correct Answer - C | |
| 177. |
Choose the false statement.A. Scalar product and vector product obey commutative lawB. Scalar product does not obey distributive law where as vector product obeys commutative lawC. Scalar product and vector product obey associative lawD. All the above |
| Answer» Correct Answer - D | |
| 178. |
Distributive law is obeyed byA. scalar productB. vector productC. bothD. none |
| Answer» Correct Answer - C | |
| 179. |
A force `F=-k(y hati + x hatj)` (where k is a positive constant) acts on a particle moving in the `x-y` plane. Starting from the origin, the particle is taken along the positive x-axis to the point `(a,0)` and then parallel to the y-axis to the point `(a,a)`. The total work done by the force F on the particle is (a) `-2ka^(2)` , (b) `2ka^(2) , (c)`-ka^(2)` , (d) `ka^(2)` |
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Answer» For motion of the particle from (0,0)to(a,0), `vec(F)=-K(0hat(i)+ahat(j))rArrvec(F)=-Kaj` Displacement, `vec(r )=(ahat(i)+0hat(j))-(0hat(i)+0hat(j))=ahat(i)` so work done from (0,0)to(a,0) is given by `W=vec(F).vec(r )=-Kahat(j).ahat(i)=0` For motion (a,0) to (a,a) `vec(F)= -K(ahat(i)+hat(j))` and displacement, `vec(r )=(ahat(i)+ahat(j))-(ahat(i)+0hat(j))=ahat(j)` So work done from `(a,0)` to`(a,a)` `W=vec(F).vec(r)= -K(ahat(i)+ahat(j)).ahat(j)= -Ka^(2)` So total work done `= -Ka^(2)` |
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| 180. |
A particle is moving eastwards with a velocity of ` 5 ms_(-1)`. In `10 seconds` the velocity changes to `5 ms^(-1)` northwards. The average acceleration in this time isA. ZeroB. `1/(sqrt(2))m//s^(2) N-W`C. `1/(sqrt(2))m//s^(2)N-E`D. `1/(sqrt(2))m//s^(2) S-W` |
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Answer» Correct Answer - B `Deltav= 2v sin ((theta)/(2))= 2xx5xxsin 45^(@)= (10)/(sqrt(2))` `:. a=(Deltav)/(Deltat)= (10sqrt(2))/(10)= 1/(sqrt(2))m//s^(2)` |
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| 181. |
If `vec(A) = 2 I + j - k , vec(B) = I + 2 j + 3 k` , and `vec(C ) = 6 i - 2 j - 6 k`,then the angle between `(vec(A) + vec(B))` and `vec(C )` will beA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - D `vec(A) + vec(B) = vec(D) = 3 i + 3j + 2k` `vec(C ) = 6 i - 2j - 6k` `vec(D) . Vec(C ) = 18 - 6 - 12 = 0` `theta = 90^(@)` |
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| 182. |
The intantaneous coordinates of a particle are ` x= (8t-1)m` and `y=(4t^(2))m`. Calculate (i) the average velocity of the particle during time interval from `t =2s ` to `t=4s` (ii) the instantaneous velocity of the particle at `t = 2s`. |
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Answer» (i) Given that `x(t) = (8t-1)` m and `y=(4t^(2))m` `r(t) = (8t- 1)hati + (4 t^(2) )hatj" ".....(i)` At `t=2s," "vecr_(1) = 15 hati + 16 hatj` At `t= 4s," "vecr_(2) = 31 hati + 64 hatj` `because` Displacement, `vecr_(2) - vecr_(1) = (31 hati + 64 hatj)- (15 hati + 16hatj)` `" "=16 hati + 48 hatj` `therefore` Average velocity `vecv_(av) =(vecr_(2) -vecr_(1))/(vect_(2) + vect_(1))` `" "(16hati + 48hatj)/(4-2)` `" "= (8hati + 24hatj) ms^(-1)` (ii) Differentianting Eqn. (i) w.r.t. time, `" "vecv= (dvecr)/(dt)= 8hati + 8thatj` At `t=2 2s, vecv = 8 hati + 16 hatj` |
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| 183. |
The angle between two vectors given by `6bari+6barj-3bark` and `7bari+4barj+4bark` isA. `cos^(-1)(1/(sqrt(3)))`B. `cos^(-1)(5/(sqrt(3)))`C. `sin^(-1)(2/(sqrt(3)))`D. `sin^(-1)((sqrt(5))/3)` |
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Answer» Correct Answer - D |
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| 184. |
To go from town A to town B a plane must fly about 1780 km at an angle of `30^(@)` west of north. How far West of A is B?A. 1542kmB. 1452kmC. 1254kmD. 890 km |
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Answer» Correct Answer - D `x=d sin theta` |
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| 185. |
Calculate the values of (i) `hatj. (2hati - 3hatj +hatk)` and (ii) `(2hati - hatj) (3hati + hatk)` |
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Answer» Correct Answer - (i) -3 (ii)6 |
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| 186. |
If the projections ofvector ` vec a`on `x`-, `y`- and `z`-axes are 2, 1 and 2 units,respectively, find the angle at which vector ` vec a`is inclined to the `z`-axis. |
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Answer» Since projections of vector `veca` on x-, y- and z-axes are 2, 1 and 2 units, respectively, we have `" ""Vector "veca=2hati+hatj+2hatk` `" "|veca|=sqrt(2^(2)+1^(2)+2^(2))=3` Then `cosgamma=(2)/(3)` (where `gamma` is the angle of vector `veca` with the z-axis), i.e., `" "gamma=cos^(-1)""(2)/(3)` |
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| 187. |
`A B C D E`is pentagon,prove that` vec A B`+ ` vec B C`+ ` vec C D`+ ` vec D E+ vec E A`= ` vec0`` vec A B+ vec A E+ vec B C+ vec D C+ vec E D+ vec A C=3 vec A C` |
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Answer» `vecR=vec(AB)+vec(AE)+vec(BC)+vec(DC)+vec(ED)+vec(AC)` `" "=(vec(AB)+vec(B C))+(vec(AE)+vec(ED)+vec(DC))+vec(AC)` `" "=vec(AC)+vec(AC)+vec(AC)=3vec(AC)` |
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| 188. |
Assertion: Multiplying any vector by an scalar is meaningful operatons. Reason: In uniform motion spedd remains constant.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both aseertion and reason are false. |
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Answer» Correct Answer - B We can multiply any vector by any scalar. For example, in equation `vec(F)= mvec(v)` mass is a scalar quantity, but acceleration is a vector quantity. |
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| 189. |
The component of vector `A= 2hat(i)+3hat(j)` along the vector `hat(i)+hat(j)` isA. `5/(sqrt(2))`B. `10sqrt(2)`C. `5sqrt(2)`D. 5 |
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Answer» Correct Answer - A `(vec(A).vec(B))/(|hat(i)+hat(j)|)=((2hat(i)+3hat(j))(hat(i)+hat(j)))/(sqrt(2))=(2+3)/(sqrt(2))= 5/(sqrt(2))` |
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| 190. |
The component of vector `A= 2hat(i)+3hat(j)` along the vector `hat(i)+hat(j)` isA. `5`B. `5//sqrt(2)`C. `10sqrt(2)`D. `2 sqrt(5)` |
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Answer» Correct Answer - B Scalar component of `vec(A) along vec(B)` is `|vec(A)| cos theta = |vec(A)| (vec(A). vec(B))/(|vec(A)| |vec(B)|) = vec(A) . (vec(B))/(|vec(B)|) = vec(A) . hat(B)` ` hat(B) = (vec(B))/(|vec(B)|) = ( i + j)/(sqrt( 1 + 1)) = (1)/(sqrt(2)) ( i + j)` `vec(A) . hat(B) = ( 2 i + 3 j) . (1)/(sqrt(2))( i + j) = ( 2 + 3)/(sqrt(2)) = (5)/(sqrt(2))` |
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| 191. |
Find the unit vector perpendicular to ltbr. `vecA=3hati+2hatj-hatk` and `vecB=hati-hatj+hatk` |
| Answer» `hateta=(vecAxxvecB)/(|vecAxxvecB|)=(hati-4hatj-5hatk)/(sqrt(1+16+25))=(hati-4hatj-5hatk)/(sqrt(42))` | |
| 192. |
A vector perpendicular to both the vector `2hati -3hatj and 3hati - 2hatj ` isA. `hatj + 5hatk`B. `hatj -5hatk`C. `6hatk`D. `hati +hatj+ hatk` |
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Answer» Correct Answer - C (c ) For vector to be perpenidicular` A.B=0` `therfore"the appropriate vector will be "6hatk` |
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| 193. |
Consider a force vector `F=hati+ hatj + hatk ` Anorther vectoer perpendicular of F isA. `4hati +3hatj`B. `6hati`C. `2hati -hatj -hatk`D. `3hati - 4hatj` |
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Answer» Correct Answer - C ( c) Forvector to be perpendicular ,A.B=0 `therfore F=2hati-hatj-hatk` |
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| 194. |
If two non-parallel vectors `vecA` and `vecB` are equal in magntiude, them vectors `(vecA - vecB)` and `(vecA + vecB)` will be :A. parallel to each otherB. parallel but oppositely directedC. perpendicular to each otherD. inclined at an angle `theta` always less than `90^(@)` |
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Answer» Correct Answer - C `(vecA- vecB) (vecA+ vecB)=0` `A^(2) - B^(2) =0` `cos theta =(|vecA|^(2) - |vecB|^(2))/(|A||B|) ` `rArr" "(0)/(|A||B|) =0` `theta = (pi)/(2)` |
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| 195. |
If `a+b+c=0, "then "axxb ` is equal toA. `bxxc`B. `cxxb`C. `axxc`D. None of these |
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Answer» Correct Answer - A (a) We have a+b+c=0 `therfore a+C=-b` or`(a+c)xxb=-bxxb=0` `or(axxb)+(cxxb)=0` `oraxxb=bxxc` |
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| 196. |
Resultant of two vector of equal magnitude A isA. `sqrt(3)Aat 60^(@)`B. `sqrt(2)Aat 90^(@)`C. `2Aat 120^(@)`D. `Aat 180^(@)` |
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Answer» Correct Answer - A,B (a,b) Resultant of two vector `A_(R)=2acos(theta)/(2)` |
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| 197. |
If the vectors `(hati+ hatj + hatk)` and `3 hati` from two sides of a triangle, then area of triangle is :A. `sqrt(3)`B. `2sqrt(3)`C. `3//sqrt(2)`D. `3sqrt(2)` |
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Answer» Correct Answer - C `vecA xx vecB =|{:(hati,hatj,hatk),(1, 1, 1),(3,0,0):}|= 3(hati - hatk)` Area of `Delta = (1)/(2)|vecA xx vecB| =(3)/(2)|hatj - hatk| =(3)/(2) sqrt(2) =(3)/(sqrt(2)` |
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| 198. |
If `ahati +bhatj` is a unit vector and it iws perpendicular to `hati +hatj` , then value of a and b isA. 1,0B. `-2,0`C. `0.5,-0.5`D. None of these |
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Answer» Correct Answer - D (d) `(ahati+bhatj).9hati+hatj_=0or a+b=0` "Futher"sqrt(a^(2)+b^(2))=1` Or`a^(2)+b^(2)=1` `"solvingEqs,(i)and(ii)we get"` `a=+-(1)/(sqrt(2))` |
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| 199. |
Two vectors `vec(a)` and `vec(b)` are such that `|vec(a)+vec(b)|=|vec(a)-vec(b)|`. What is the angle between `vec(a)` and `vec(b)`?A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - D Let `theta` be the angle between the vectors `vec(A)` and `vec(B)`. Then `|vec(A)+vec(B)|= sqrt(A^(2)+B^(2)+2AB cos theta)` `|vec(A)-vec(B)|= sqrt(A^(2)+B^(2)-2AB cos theta)` `|vec(A)+vec(B)|=|vec(A)-vec(B)|` `:. sqrt(A^(2)+B^(2)+2ABcos theta)=sqrt(A^(2)+B^(2)-2AB cos theta)` Squarung both side, we get `A^(2)+B^(2)+2AB cos theta= A^(2)+B^(2)-2AB cos theta` `4AB cos theta=0` `:. cos theta=0 or theta= (pi)/2` |
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| 200. |
Following forces start acting on a particle at rest at the origin of the co-ordiante system simultaneously `vecF_1=4hati-4hatj+5hatk`,`vacF_2=5hati+8hatj+6hatk`,`vecF_3=-3hati+4hatj-7hatk` and `vecF_4=2hati-3hatj-2hatk` then the particle will moveA. In `x-y` planeB. In `y-z` planeC. In `x-z` planeD. Along `x`-axis |
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Answer» Correct Answer - A |
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