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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. |
A coins kept on a horizontal rotating disc has its centre at a distance of 0.25 m from the axis of rotation of the disc .If `mu is 0.2`, then the angular velocity of the disc at which the coion will slip off, `(g=9.8m//s^(2))`A. `3.8 rad//s`B. `2.8 rad//s`C. `4.8 rad//s`D. `5.8 rad//s` |
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Answer» Correct Answer - B `v=sqrt(mu g r )` `therefore r omega =sqrt mu g r ` `therefore omega = sqrt (mu g)/(r)` |
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| 152. |
Three balls each of mass 1 kg are attached with three strings each of length 1 m as shown in figure. They are rotated in a horizontal circle with angular velocity `omega=4"rad s"^(-1)` about point O. Match the following columns. |
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Answer» Correct Answer - (A)P,(B)R,(C),QT |
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| 153. |
Three balls each of mass 1 kg are attached with three strings each of length 1 m as shown in figure. They are rotated in a horizontal circle with angular velocity `omega=4"rad s"^(-1)` about point O. Match the following columns. |
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Answer» Correct Answer - A::B::C `T_(3)=(1)(3)(4)^(2)=48N" "(becauseF=mRomega^(2))` `T_(2)-T_(3)=(1)(2)(4)^(2)` `therefore" "T_(1)=80NrArrT_(1)-T_(2)=(1)(1)(4)^(2)` `therefore" "T_(1)=96 N` |
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| 154. |
Four point size metal spheres each of mass 1 kg are placed on a turn table and are connected by four strings of equare. If the spheres are rotated with an angular velocity `(1)/(pi)` rps, the tention in the connecting strings isA. 4 NB. 2 NC. 1 ND. 3 N |
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Answer» Correct Answer - B If T is the tension in each string, `sqrt2T=mr omega^(2)` `but , r=(l)/(sqrt2)=(1)/(sqrt2) m ` ` T = (mromega^(2))/(sqrt2) ((1)((1)/(sqrt2)) (2pi (1)/(pi))^(2))/(sqrt2)` `=(4)/(2)=2N`. |
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| 155. |
When the angular velocity of a uniformly rotating body has increased thrice, the resultant of forces applied to it increases by 60 N. Find the accelerations of the body in the two cases. The mass of the body,m = 3 kg.A. `2.5ms^(-2),7.5ms^(-2)`B. `7.5ms^(-2),22.5ms^(-2)`C. `5 ms^(-2),45ms^(-2)`D. `2.5ms^(-2),22.5ms^(-2)` |
| Answer» Correct Answer - D | |
| 156. |
A particle moves in a particle of radius `0.5 m` at a speed that uniformly increases. Find the angular acceleration of particle if its speed changes from `2.0m//s` to `4.0 m//s` in `4.0s` |
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Answer» Correct Answer - A::B::D The tangential acceleration of the particle is `a_(t)=(dv)/(dt) =(4.0-2.0)/4.0=0.5 m//s^(2)` The angular acceleration is `alpha=(a_(t))/r=0.5/0.5=1 rad//s^(2)` |
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| 157. |
Two identical balls 1 and 2 are tied to two strings as shown in figure. They are rotated about point O. Ball 1 is observed from ball 2, centrifugal force on ball 1 is `F_(1)`. Similarly, ball 2 is observed from ball 1 and centrifugal force on ball 2 is `F_(2)`, then A. `|F_(1)|=|F_(2)|=0`B. `|F_(1)|gt|F_(2)|`C. `|F_(1)|=|F_(2)|ne 0`D. `F_(1) and F_(2)` are antiparallel |
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Answer» Correct Answer - B |
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| 158. |
A article moves in a circle of radius 20 cm with linear speedof 10 m/s. Find the angular velocity |
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Answer» The angular velocity is `omega=v/r=(10m/s)/(20cm)=50 rad/s. |
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| 159. |
A particle travels in a circle of radius 20 cm at a speed thast uniformly increases. If the speed changes from 5.0 m/s to 6.0 m/s in 2.0s, find the angular aceleration. |
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Answer» The tangential acceleration is gilven by ` a_t=(dv)/(dt)=(v_2-v_1)/(t_2-t_1)` `=(6.0-5.0)/2.0m/s^2=0.5m/s^2` The angula accelerationis `alpha =a_t/r` `=(0.5m/s^2)/(20 cm)=2.5rads^2` |
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| 160. |
A body moves on a horizontal circular road of radius `r`, with a tangential acceleration `a_(t)`. The coefficient of friction between the body and the road surface Is `mu`. It begins to slip when its speed is `v`. (i) `v^(2)=murg` (ii) `mug=(v^(4)/(r^92))+a_(t))` (iii) `mu^(2)g^(2)=(v^(4)/(r^(2)+a_(t)^(2))` (iv) The force of friction makes an angle `tan^(-1)(v^(2)//a_(t)r)` with the direction of motion at the point of slipping.A. `v^(2)=murg`B. `mug=v^(2)/r+a_(T)`C. `mu^(2)g^(2)=v^(4)/r^(2)+a_(T)^(2)`D. The force of friction makes an angle `tan^(-1)((v^(2))/(a_(T)xxr))` with direction of motion of point of slipping. |
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Answer» Correct Answer - C::D At time of slipping `f=mumg` `f cos theta=ma_(T), f^(2)=(ma_(T))+((mv^(2))/r)` `(mumg)^(2)=(ma_(T))^(2)+((mv^(2))/r)^(2)` `rArr mu^(2)g^(2)=a_(T)^(2)+((v^(4))/r^(2))`,Also `tan theta=v^(2)/(a_(T)r)` |
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| 161. |
If `a_(r )` and `a_(t)` respresent radial and tangential acceleration, the motion of a particle will be circular isA. `a_(r )=0` and `a_(t)=0`B. `a_(r)=0` but `a_(t)!=0`C. `a_(r )!=0` but `a_(t)=0`D. `a_(r )!=0` and `a_(t)!=0` |
| Answer» Correct Answer - C | |
| 162. |
A particle mass `m` begins to slide down a fixed smooth sphere from the top of its vertical diameter. Calculate its tangential acceleration, radial acceleration and total acceleration when it breaks off.A. `(2g)/(3)`B. `(sqrt(5)g)/(3)`C. `g`D. `(g)/(3)` |
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Answer» Correct Answer - B |
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| 163. |
A stone tied to a string is rotated in a circle. If the string is cut, the stone flies away from the circle becauseA. a centrifugal force acts on the stoneB. a centrifugal force acts on the stoneC. of its ineratia of motionD. dreaction of the centripetal |
| Answer» Correct Answer - C | |
| 164. |
A particle of mass 20 kg executing uniform circular motion on a path of radius 4.5 m , if the `30kg m//s` is the magnitude of licear momentum . Then the radian force acting on the particle isA. 40 NB. 20 NC. 30 ND. 10 N |
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Answer» Correct Answer - D `F=(P^(2))/(rm)=(30xx30)/(4.5xx20)=10N` |
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| 165. |
A point starts from rest and moves along a circular path with a constant tangential acceleration. After one rotation, the ratio of its radial acceleration to its tangential acceleration will be equal toA. 1B. `2pi`C. `(1)/(2)pi`D. `4pi` |
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Answer» Correct Answer - D `(a_(n))/(a_(t))=(v^(2)//R)/(a)" "("let a"_(t)=a)` Here, `v^(2)=2al=2a(2piR)=4piR`. Therefore, the ratio is `(4pi)/(1)`. |
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| 166. |
A paticle of mass `m` is executing uniform circular motion on a path of radius `r`. If `p` is the magnitude of its linear momentum, then the radial force acting on the particle isA. pmrB. `(rm)/(p)`C. `(mp^(2))/(r)`D. `(p^(2))/(rm)` |
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Answer» Correct Answer - D Radial force `=(mv^(2))/(r)=(m)/(r)((p)/(m))^(2)=(p^(2))/(mr)` |
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| 167. |
A motorcycle is travelling on a curved track of radius 500 m. If the coefficient of friction between road and tyres is 0.5, the speed avoiding skidding will beA. `500 m//s`B. `250 m//s`C. `50 m//s`D. `19 m//s` |
| Answer» Correct Answer - C | |
| 168. |
A block sliding down on a rough `45^(@)` inclined planes has half the velocity it would have been, the inclined plane is smooth. The coefficient of sliding friction between the the block and the inclined plane isA. `1/4`B. `3/4`C. `1/(2sqrt2)`D. `1/sqrt2` |
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Answer» Correct Answer - B `v^(2)-u^(2)=2as` `V_(R)=sqrt(2gl(sin theta-mu_(k) cos theta)),V_(S)=sqrt(2 gl sin theta)` |
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| 169. |
A turn of radius 600 m is banked for a vehicle of mass 200 kg going with a speed of `180kmh^(-1)` . Determine the banking angle of its path. |
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Answer» The turn is banked for speed, `v=180"kmh"^(-1)=180xx(5)/(18)"ms"^(-1)=50ms^(-1)` and radius r = 600m `because" "tantheta=(v^(2))/(rg)=(50xx50)/(600xx10)` `rArr" "tantheta=(25)/(60)=0.417` `rArr" Banking angle "theta=tan^(-1)(0.417)` `rArr" "theta=22.6^(@)` |
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| 170. |
A body rest on the top of a hemisphere of radius R. What will be the least horizontal velocity imparted to it , if it has to leave the hemisphere without sliding down?A. `sqrt(2gR)`B. `sqrt(5gR)`C. `sqrt(gR)`D. `sqrt(3gR)` |
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Answer» Correct Answer - C `v=sqrt(gR)` |
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| 171. |
A small coin is kept at the rim of a horizontal circular disc which is set into rotation about verticle axis passing through its centre. If radius of the disc is `5 cm and mu_(s)=0.25`, then the angular speed at which the coin will just slip off atA. ` 5 rad//s`B. `7 rad//s`C. `10 rad//s`D. `4.9 rad//s` |
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Answer» Correct Answer - B `v=sqrtmu r g and omega =(v)/(r)` |
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| 172. |
A ball of mass 100 g released down an inclined plane describes a circle of radius 10 cm in the vertical plane on reaching the bottom of the inclined plane . The minimum height of the incline isA. `25 cm `B. `15 cm`C. `30 cm`D. `10 cm` |
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Answer» Correct Answer - A `h=(5r)/(2)=(5(10))/(2)=25 cm` |
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| 173. |
A bucket tied at the end of a `1.6m` long string is whirled in a verticle circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill, when the bucket is at the highest position `(Takeg=10m//s^(2))`A. 2m/secB. 10 m/secC. 4 m/secD. 5 m/sec |
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Answer» Correct Answer - C For water from the bucket does not spill, N`ge`0 at highest point, for `N ge0`, so minimum constant speed. `v=sqrt(rg)=sqrt(1.6xx10)=4 m//s` |
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| 174. |
An object of mass 2 kg is whirled rround in a verticle circle of radius 1m with a constant speed of `4 m//s` .Then the maximum tension in the string is `(g=10 m//s^(2))`A. 32 NB. 52 NC. 72 ND. 92 N |
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Answer» Correct Answer - B `T_(max)=(mv^(2))/(r)+mg` `=(2xx16)/(1)+2xx10=32+20` `=52 N` |
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| 175. |
A mass of 2 kg tied to a string 1 m length is rotate in a verticle circle with a uniform speed of `4 m//s`. The tension in the string will be 52 N, when the mass is at `(g=10 m//s^(2))`A. bottomB. highest pointC. midwayD. horizontal position |
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Answer» Correct Answer - A `T=(mv^(2))/(r)+m g cos theta ` `therefore mg cos theta =T-(mv^(2))/(r)` `=52-(2xx16)/(1)` ` cos theta =(20)/(mg)=(20)/(2xx10)=1` `cos theta =1` `theta =cos^(-1) (0)=0`. |
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| 176. |
A body of mass 0.5 kg is whirled in a verticle circle an angular frequancy of `10 rad //s`.If the radius of the circle is 0.5 m . What is the tension in the string when the body is at the top of the circle? `(Given g= 10 m//s^(2))`A. 10NB. 20 NC. 30 ND. 40 N |
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Answer» Correct Answer - B `T =mr omega ^(2)-mg` `=0.5xx0.5xx100-0.5xx10` `=25-5` `20N` |
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| 177. |
A stone of mass 1 kg tied at the end of a string of length 1 m and is whirled in a verticle circle at a constant speed of `3 ms^(-1)`. The tension in the string will be 19 N when the stone is `(g=10 ms^(-1))`A. at the top most point on the vertical circleB. at the bottom most point on the vertical circleC. half way downD. making an angle `30^(@)` with the vertical |
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Answer» Correct Answer - B `T=(mv^(2))/(r)+mg cos theta` `mg cos theta=T-(mv^(2))/(r)=19-(1xx9)/(1)=19-9` `mg cos theta =10` `therefore cos theta =(10)/(mg)=(10)/(1xx10)` `cos theta =1` `theta= cos ^(-1)(1)=0` At the bottom |
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| 178. |
A `1kg` stone at the end of `1m` long string is whirled in a vertical circle at a constant speed of `4m//s`. The tension in the string is `6N`, when the stone is at `(g=10m//s^(2))`A. top of the circleB. bottom of the circleC. half way downD. non of the above |
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Answer» Correct Answer - A `mg=1xx10=10N,(mv^(2))/(r)=(1xx(4)^(2))/(1)=16` Tension at the top of circle `=mv^(2)/(r)=mg=6N` Tension at the bottom of circle `(mv^(2))/(r)+mg=26N` |
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| 179. |
A stone is thrown horizontally with a velocity of `10m//sec`. Find the radius of curvature of it’s trajectory at the end of `3 s` after motion began. `(g=10m//s^(2))`A. `10sqrt10m`B. `100sqrt10m`C. `sqrt10m`D. `100 m` |
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Answer» Correct Answer - A Radius of curvature `r=v_(t)^(2)/(g sin alpha)` `V=sqrt(v_(x)^(2)+v_(y)^(2))=10sqrt10` `tan alpha=v_(x)/v_(y)=1/3, sin alpha=1/sqrt10` |
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| 180. |
A particle is projected with a speed u at angle `theta` with the horizontal. Consider a small part of its path ner the highest position and take it approximately to be a circular arc. What is the rdius of this circle? This radius is called the adius of curvature of the curve at the point. |
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Answer» Correct Answer - A::B::C Particle is projected with speed u at an angle `theta`. At the highest point the veticasl component of velocity is 0. So at the point, velocity`=ucostheta` Centripetal force `=(mu^2cos^2theta)/r` At highest point `mg=(mv^2)/r=(mu^2 cos^2theta)/r` `rarr r=(u^2cos^2theta)/g` |
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| 181. |
A stone is projected with speed u and angle of projection is `theta`. Find radius of curvature at t=0.A. `(u^(2) cos^(2) theta)/g`B. `(u^(2))/(g sin theta)`C. `(u^(2))/(g cos theta)`D. `(u^(2) sin^(2) theta)/g` |
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Answer» Correct Answer - C At t=0 `a_(bot)=g cos theta`, `R=(v^(2))/(a_(bot))=(u^(2))/(g cos theta)` |
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| 182. |
A particle is projected with a speed u at angle `theta` with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circle? This radius is called the adius of curvature of the curve at the point. |
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Answer» Correct Answer - A::B::C `a=(v^(2))/(R) implies R=(v^(2))/(a)=((ucostheta)^(2))/(g)` |
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| 183. |
A body of mass `2kg` is moving with a velocity of `vecu=3hati+4hatj m//s`.A steady force `vecF=hati-2hatj N` begins to act on it. After four second, the body will be moving along.A. `X`-axis with a velocity of `2 m//s`B. `Y`-axis with a velocity of `5 m//s`C. `X`-axis with a velocity of `5 m//s`D. `Y`-axis with a velocity of `2 m//s` |
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Answer» Correct Answer - C `v=u+at,F=ma` |
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| 184. |
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m . Give the magnitude and direction of (a) the force on the `7^(th)` coin (counted from the bottom due to all the coins on its top . (b) the force on the `7^(th)` coin by the eigth coin. (c ) the reaction of the `6^(th)` coin one th `7^(th)` coin .A. `0.3 N` downwardsB. `0.3 N` upwardsC. `0.7 N` downwardsD. `0.7 N` upwards |
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Answer» Correct Answer - A Force on the seventh coin = weight of the three coins above it. `F=3mg` |
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| 185. |
What is the radius of curvature of the parabola traced out by the projectile.Projected with a speed `u=sqrt30` at angle `theta=60^(@)` with the horizontal at a point where the particle velocity makes an angle `theta//2` with the horizontal ? |
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Answer» Correct Answer - A As velocity along horizontal remains constant `U cos theta=V cos (theta/2)` Radius of curvature `r=v^(2)/a_(T)=(v^(2)cos^(2)theta)/(g cos^(2)(theta//2)` |
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| 186. |
In a typical projectile motion, tangential acceleration at the topmost point P of the trajectory is A. gB. `gcostheta`C. 0D. None of these |
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Answer» Correct Answer - C At highest point of a projectile, tangential compoment of acceleration is zero as acceleration is vertically downwards (g) and velocity is horizontal or the angle between acceleration and velocity is `90^(@)`. |
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| 187. |
A small ball descibes a horozontal circle on the smooth inner surface of a conical funnel. If the height of the plane of the circle above the vertex be 10 cm what is the speed of the particle?A. `2 m//s`B. `1m//s`C. `4 m//s`D. `10 m//s` |
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Answer» Correct Answer - B `v=sqrt((rg)/(tan theta))` |
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| 188. |
Cement, sand and seree are dropped in rotating cylidrical drum to make concrete mixture. If rotating speed of drum is very high then contents are attached to wall of drum and mixture is not formed correctly. If radius of drum is 1.25 m and its axis is horizontal, then the required maximum rotating speed to make good mixture in rpm is -A. 8B. 0.4C. 1.3D. 27 |
| Answer» Correct Answer - D | |
| 189. |
A block of metal weighing 2kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of `1kg s^(-1)` and at a speed of `5ms^(_1)`. The initial acceleration of the block is A. `2.5 m//s^(2)`B. `5 m//s^(2)`C. `10 m//s^(2)`D. `20 m//s^(2)` |
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Answer» Correct Answer - A `F=(dp)/(dt),F=v(dm)/(dt),a=F/m` |
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| 190. |
Find the vector perpendicular to the plne determined by the points P(1,-2,2), Q(2,0,-1 ) and R(0,2,1) |
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Answer» `bar(PQ)=("Position vector of Q")-("Position vector of P")` `=(2hati-hatk)-(hati-hatj+2hatk)=hati+hatj=3hatk` `"Similarly", bar(P)R=(hat2j+bark)-(hati-hatj+2hatk)=-hati+3hatj+bar(k)` `therefore |bar(PQ)xxbar(PR)|=sqrt(8^(2)+4^(2)+4^(2))=4sqrt6` Required unit vectors `=pm(1)/(4sqrt6)(8hati+4hatj+4hatk)=pm(1)/(sqrt6)(2hati+hatj+8hatk)` |
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| 191. |
In circular motion, what are the possible values (zero, positive or negative) of the following: (a) `omega.v` (b) `v.a`, (c) omega.alpha` |
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Answer» (a) `v` lies in the plane of circle and `omega` is always perpendicular to this plane. `:. Omega bot omega` Hence,. `omega.v` is always zero. (b) `v` and `a` both lie in the plane of circle and the angle between these two vectors may be acute (when speed is increasing) obtuse (when speed is decreasing) or `90^(@)` (when speed is consants) (c) `omega` and `alpha` either parallel `(theta=0^(@))` between `omega` and `alpha`) or antiparallel `(theta=180^(@))`. in uniform circular motion, `alpha` has zero magnitude. hence,. `omega. alpha` may be positive, negative or zero. |
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| 192. |
A point moves along a circle having a radius `20cm` with a constant tangential acceleration `5cm//s^(2)` . How much time is needed after motion begins for the normal acceleration of the point to be equal to tangential acceleration?A. `1s`B. `2s`C. `3s`D. `4s` |
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Answer» Correct Answer - B `(v^(2))/(R)=a_(t)=a` (Here, `a_(t)=`asay) or `((at)^(2))/(R)=a` :. `t=sqrt((R)/(a))=sqrt((20)/(5))=2s` |
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| 193. |
A car runs east to west and another car B of the same mass runs from west to east at the same path along the equator. A precess the track with a force `N_(1)` and B presses the track with a force `N_(2)`. ThenA. `F_1gtF_2`B. `F_1ltF_2`C. `F_1=F_2`D. `the information is insufficietn to find the relation between `F_1 and F_2`. |
| Answer» Correct Answer - A | |
| 194. |
A car moves at a constant speed on a road as shown in figure. The normal force by the road on the car is `N_A and N_B` when it is at the points A and B respectively.A. `N_A=N_B`B. `N_AgtN_B`C. `N_AltN_B`D. insufficient informtion to decide the relation of `N_A and N_B` |
| Answer» Correct Answer - C | |
| 195. |
When a particle moves in a circle with a uniform speedA. its velocity and acceleration re both constantB. its velocilty is constant but the acceleration changesC. its acceleration is constant but the velocity changesD. its velocity and acceleration both change. |
| Answer» Correct Answer - D | |
| 196. |
A small particle of mass `m` attached with a light inextensible thread of length `L` is moving in a verical circle. In the given case particle is moving in complete vertical circle and ratio of its maximum to minimum velocity is `2:1` . Kinetic energy of the particle at the lower most position is A. `(4mgL)/(3)`B. `2mgL`C. `(8mgL)/(3)`D. `(2mgL)/(3)` |
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Answer» Correct Answer - C `u_(max)=2u_(min)=4sqrt((gL)/(3))` :. `K_(max)=(1)/(2)mu_(max)^(2)=(8mgL)/(3)` |
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| 197. |
A small particle of mass `m` attached with a light inextensible thread of length `L` is moving in a verical circle. In the given case particle is moving in complete vertical circle and ratio of its maximum to minimum velocity is `2:1` . Minimum velocity of the particle is A. `4sqrt((gL)/(3))`B. `2sqrt((gL)/(3))`C. `sqrt((gL)/(3))`D. `3sqrt((gL)/(3))` |
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Answer» Correct Answer - B Maximum velocity is at bottommost point and minimum velocity is at tomost point. `sqrt(u_(min)^(2)+2g(2L))/(u_(min))=(2)/(1)` On solving, we get `u_(min)=2sqrt((gL)/(3))` |
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| 198. |
A satellite of the earth is revolving in a circular orbit with a uniform speed `v`. If the gravitational force suddenly disappears, the satellite willA. continue to move the velocity v along the original orbitB. move with a velocity v, tangentially to the original orbitC. fall down with increasing velocityD. ultimately come to rest somewhere on the origenal orbit |
| Answer» Correct Answer - B | |
| 199. |
Centripetal force is a real force because its origing canA. not be explainedB. be explainedC. lies in revolving changesD. be at centre of mass of a body |
| Answer» Correct Answer - B | |
| 200. |
If the earth revolves round the sun in a circular orbit, then the necessary entripetal force will be provided byA. weight of the earthB. acceleration due to gravityC. gravitational force of attaction between the sun and the earthD. presence of atmosphere around the earth |
| Answer» Correct Answer - C | |