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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.
| 1151. |
A small block of mass `M` moves on a frictionless surface of an inclined plane, as shown in the figure. The angle of the incline suddenly changes from `60^(@)` to `30^(@)` at point `B`. The block is many at rest at `A`. Assume that collisions between the block id the incline are totally inelastic. If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the blocks at point `B`, immediately after it strikes the second incline is A. `sqrt30` m/sB. `sqrt15` m/sC. 0D. `-sqrt(15)` m/s |
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Answer» Correct Answer - C The velocity of the block coming down from the incline AB makes an angle `30^@` with the incline BC. If the block collides with the incline BC elastically, the angle of velocity of the block after collision with the incline shall be `30^@` Hence just after collision with incline BC the velocity of block shall be horizontal. So immediately after the block strikes the second incline, its vertical component of velocity will be zero. |
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| 1152. |
A car of mass 1000kg accelerates uniformly from rest to a velocity of `54km//h` in 5 seconds. Calculate (i) its acceleration (ii) its gain in KE (iii) average power of the engine during this period. |
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Answer» Here, `m=1000kg, u=0` `u=54km//h=15m//s` `t=5s, a=?, KE=?, Power =?` From `v=u+at, a=(v-u)/(t)=(15-0)/(5)=3m//s^(2)` Gain in KE `=(1)/(2)mv^(2)-0=(1)/(2)xx1000(15)^(2)` `=1.125xx10^(5)J` Power `= ("work")/("time")=(KE)/(t)=(1.125xx10^(5))/(5)` `=22500watt` |
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| 1153. |
How high must a body be lifted so that it gain P.E. equal to its KE while moving with a velocity of `30m//s`?Take `g=10m//s^(2)` |
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Answer» Here, `h=?v=30m//s, g=10m//s^(2)` If m is mass of the body, then as `P.E. =K.E. :. mgh=(1)/(2)mv^(2), h=(v^(2))/(2g)=(30xx30)/(2xx10)=45m` |
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| 1154. |
A stone of mass 2 kg is projected upwards with KE of 98 J. The height at which the KE of the body becomes half its original value, is given by (take, `g=9.8ms^(-2)`)A. 5 mB. 2.5 mC. 1.5 mD. 0.5 m |
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Answer» Correct Answer - B (b) At this height, half is PE `:. Mgh=(98)/(2)" or " 2xx9.8xxh=49" or " h=2.5 m` |
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| 1155. |
A large mass `M` and a small mass `m` hang at the two ends of string passing through a smooth tube. The mass `m` moves around in a circular path in a horizontal plane. The length of the string from the mass `m` to the top of the tube of the tube is `l` and `theta` is the angle this length makes with the vertical. What should be the frequency of rotation of the mass m so that M remains stationary ?A. `1/2pisqrt(Mg)/(ml)`B. `1/2pisqrt((ml)/(Mg))`C. `1/2pi sqrt((mg)/(Ml))`D. `1/2pisqrt((mM)/(gl))` |
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Answer» Correct Answer - A |
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| 1156. |
At sea level, a `N_(2)`, molecule in air has an average translational `KE=6.2xxx10^(-21)J`. Its mass is `4.7xx10^(-26)kg.` If the molecule shoots up straight without any resistance, it will risw to a height ofA. `135km`B. `13.5km`C. `1.35 km`D. `1350 km` |
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Answer» Correct Answer - B Here,`K.E.=6.2xx10^(21)J, m=4.7xx10^(-26)kg` As `KE` of molecule will get converted into its `PE` as the water level rise. `:. PE=6.2 xx10^(21)J=mgh` or `h=(6.2xx10^(-21))/(mg)=(6.2xx10^(-21))/(4.7xx10^(-26)xx98)` `~~13.5xx10^(3)m =13.5km` |
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| 1157. |
A cannon of mass 2m located at the base of an inclined plane shoots a shell of mass m in horizontal direction with velocity `v_(0)` The angle of inclination of plane is `45^(@)` and the coefficient of friction between the cannon and the plane is `0.5` The height to which cannon ascends the plane as a result of recoil is :A. `v_(0)^(2)/(2g)`B. `v_(0)^(2)/(12g)`C. `v_(0)^(2)/(6g)`D. `v_(0)^(2)/g` |
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Answer» Correct Answer - B |
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| 1158. |
A uniform chain of length `2m` is kept on a table such that a length of `60 cm` hangas freely from the adge of the table . The table . The total mass of the chain ia `4 kg` What is the work done in pulling the entire the chain the on the table ?A. 12.9JB. 6.3 JC. 3.6 JD. 2.0 J |
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Answer» Correct Answer - C Mass of length 2 m of the chain = 4 kg Mass of length 60 cm or 0.60 m of the chain `=(4xx0.60)/2=1.2 kg` Weight of the hanging part of the chain = 1.2 x 10=12 N Since the centre of gravity of the hanging part lies at its mid point, ie. 30 cm or 0.30 m `therefore` W= 12 x 0.30= 3.6 J |
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| 1159. |
A uniform chain of length `2m` is kept on a table such that a length of `60 cm` hangas freely from the adge of the table . The table . The total mass of the chain ia `4 kg` What is the work done in pulling the entire the chain the on the table ?A. `12.9 J`B. `6.3 J`C. `3.6 J`D. `2.0 J` |
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Answer» Correct Answer - C Mass of length `2m` of the chain`= 4 kg` mass of length `60 cn or 0.60 m` of the chain `= (4 xx 0.60)/(2) = 1.2 kg` Weight of the hanging part of the chain `= 1.2 xx 10` `= 12 N` Since of center gravity of the hanging part lines at its mid point, i.e. `30 cm or 0.30 = 3.6 J` |
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| 1160. |
A uniform chain of length `2m` is kept on a table such that a length of `60 cm` hangas freely from the adge of the table . The table . The total mass of the chain ia `4 kg` What is the work done in pulling the entire the chain the on the table ?A. 7.2 JB. 3.6 JC. 120 JD. 1200 J |
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Answer» Correct Answer - B |
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| 1161. |
When a spring is stretched by 2 cm, it stores 100 J of energy. If it is further stretched by 2 cm, the stored energy will be increased byA. 100 JB. 200 JC. 300 JD. 400 J |
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Answer» Correct Answer - C |
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| 1162. |
A spring with spring constant `K` when stretched through `1 cm`, the potential energy is `U`. If it stretched by `4 cm`, the potential energy will beA. 4UB. 8UC. 16 UD. 2 U |
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Answer» Correct Answer - C |
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| 1163. |
A cyclist rides along the circumference of a circular horizontal plane of radius `R`, with the friction coefficient `mu=mu_(0)(1-(r )/(R ))`, where `mu_(0)` is constant and `r` is distance from centre of plane `O`. Find the radius of the circle along which the cyclist can ride with the maximum velocity, what is this valocity? |
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Answer» Correct Answer - A::B |
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| 1164. |
A block of mass 10 kg is hanging over a smooth and light pulley through a light string and the other end of the string is pulled down by a constant force F. The kinetic energy of the block increases by 20 J in 1 s A. Tension in string is 120 NB. Tension is string is 100 NC. The work done by tension on the block is 120 J in 1 sD. The work done by the force of gravity is 100 J |
| Answer» Correct Answer - A::C | |
| 1165. |
In figure a force of magnitude `F` acts on the free end of the card. If the weight is move up alowly by a distance b. How much work is done on the weight by the rope connecting pulley and weight? A. `Fh`B. `2 Fh`C. `Fh//2`D. None of these |
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Answer» Correct Answer - C We applying force `F` therefore tension in string is `T = F`. Hence force acting om weight is `2F` upward, and displacement is `h` upward Hence `W = 2fh` (work done on weight) |
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| 1166. |
A block of mass `m` is released from the top of a mooth inclined plane of height `h` its speed at the bottom of the plane is proportion toA. `m^(0)`B. `m`C. `m^(2)`D. `m^(-1)` |
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Answer» Correct Answer - A Let `v` be the speed of the object at the bottom of the plane According to work energy therem `W = Delta K = K_(f) - K_(i)` `mgh = (1)/(2) mv^(2) - (1)/(2) mv^(2) ( because u = 0)` `mgh= (1)/(2) mv^(2) (because u = 0)` `or v = sqrt(2gh)` From the above experession it is clear the `v` is independent of the mass of an object |
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| 1167. |
A train having `60` wawgons each weight `25` tonner moving with a speed of `72 km//h`. If the force `10 N` per tonne , the power development is:A. `3 xx 10^(5) W`B. `3 xx 10^(6) W`C. `3 xx 10^(7) W`D. `3 xx 10^(4) W` |
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Answer» Correct Answer - A Net frictional force `f = 10 xx 60 xx 25 = 15,000N` Power `= f xx v = 15,000 xx 20 = 3 xx 10^(5) W` |
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| 1168. |
A lawn roller is displaced through `1 km` using a force of `200 N` In a direction making an angle of `60^(@)` with the lawn .The work down is:A. `10^(5) J`B. `10^(4) J`C. `10^(6) J`D. `10^(3) J` |
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Answer» Correct Answer - A `W = F_(s) cos theta = 200 xx 1000 xx (1)/(2) = 10^(5) J` |
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| 1169. |
A particle moves from position `3hati+2hatj-6hatk` to `14hati+13hatj+9hatk` due to a uniform force of `(4hati+hatj+3hatk)N`. If the displacement in meters then work done will beA. 100JB. 50 JC. 200 JD. 75 J |
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Answer» Correct Answer - A |
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| 1170. |
A particle moves from position `3hati+2hatj-6hatk` to `14hati+13hatj+9hatk` due to a force `vecF=(4hati+hatj+3hatk)`N. If the displacement is in centimeter then work done will beA. 1 JB. 2 JC. 3 JD. 2.5 J |
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Answer» Correct Answer - A (a) `r_(1)=3hati+2hatj-6hatk, r_(2)=14hati+13hatj+9hatk` `d=r_(2)-r_(1)=(11hati+11hatj+15hatk)cm` `F=(4hati+hatj+3hatk) N` `W=F.d=(44+11+45)xx10^(-2)=1 J` |
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| 1171. |
A particle acted upon by constant forces `4hati +hatj- 4 hatk` and `3hati + hatj - hatk` is displacment from the point `hati+ 2hatj+ hatk` to point `5hati + 4hatj +hatk`.Total work done by the forces in SI unit is :A. 20B. 40C. 50D. 30 |
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Answer» Correct Answer - B Here, `vecF_1=4hati+hatj-3hatk,vecF_2=3hati+hatj-hatk` `vecr_1=hati+2hatj+3hatk, vecr_2=5hati+4hatj+hatk` Displacement , `vecr=vecr_2-vecr_1` `=(5hati+4hatj+hatk)-(hati+2hatj+3hatk)=4hati+2hatj-2hatk` Work done by the forces, `W=[vecF_1+vecF_2].vecr` `=[(4hati+hatj-3hatk)+(3hati+hatj-hatk)].(4hati+2hatj-2hatk)` `=(7hati+2hatj-4hatk).(4hati+2hatj-2hatk)=28+4+8=40 J` |
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| 1172. |
A force acting on particle is given by `vecF=(3x^2hati+4yhatj)N`. The change in kinetic energy of particle as it moves from `(0,2m)` to `(1m,3m)` isA. `6J`B. `10J`C. `11J`D. `13J` |
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Answer» Correct Answer - C `vecF=3x^2hati+yhatj`,`dvecr=dxhati+dyhatj` `dW=vecF.dvecr=3x^2dx+4ydy` `W=int_(0)^(1)3x^2dx+int_(2)^(3)4ydy` `=|x^3|_0^1+|2y^2|_2^3` `=1+2(3^2-2^2)=11J` `W=triangleK=11J` |
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| 1173. |
A mass of 1 kg is acted upon by a single force `F=(4hati+4hatj)N`. Under this force it is displaced from (0,0) to (1m,1m). If initially the speed of the particle was 2 `ms^(-1)`, its final speed should beA. `6 ms^(-1)`B. `4.5 ms^(-1)`C. `8 ms^(-1)`D. `4 ms^(-1)` |
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Answer» Correct Answer - B (b) Given, force, `F=(4hati+4hatj) N` `r_(1)=0hati+0hatj` `r_(2)=hati+hatj` `r_(2)-r_(1)=hati+hatj` Initial speed, `v_(1)=2 ms^(-1)` From work energy theorem we have `DeltaW=DeltaK` ` implies " " F.Delta r=(1)/(2)m(v_(2)^(2)-v_(1)^(2))` `implies " "(4hati+4hatj).(hati+hatj)=(1)/(2)xx1(v_(2)^(2)-4)` `implies " "4+4=(1)/(2)(v_(2)^(2)-4)` `implies " " v_(2)^(2)=20 implies =5xx10=50 N` |
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| 1174. |
Find the work done if a particle is displaced from `J(1m,2m,3m)` to `K(2m,3m,4m)` under a force (a) `vecF=(2hati+5hatj+6hatk)N` and (b) `vecF=(2xhati+3y^2hatj+4z^3hatk)N` |
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Answer» (a) force is constant, Position vector of J, `vecr_1=hati+2hatj+3hatk` Position vector of K, `vecr_2=2hati+3hatj+4hatk` displacement vector `vecd=vecr_2=hati+hatj+hatkm` `W=vecF.vecd=(2hati+5hatj+6hatk).(hati+hatj+hatk)` `=2+5+6=13J` (b) `vecF=2xhati+3y^2hatj+4z^3hatk=F_xhati+F_yhatj+F_zhatk` force is variable . `W=intvecF.dvecr` `=int_(x1)^(x2)F_(x)dx+int(y1)^(y2)F_ydy+int_(z1)^(z2)F_zdz` `=int_(1)^(2)2xdx+int_(2)^(3)3y^2dy+int_(3)^(4)4z^3dz` `|x^2|_1^2+|y^3|_2^3+|z^4|_3^4` `={(2)^2-(1)^2}+{(3)^2-(2)^2}+{(4)^4-(3)^4}` `=(4-1)+(27-8)+(256-81)` `=3+19+175=197J` |
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| 1175. |
A body of mass 2 kg is acted upon by two forces `vecD_(1)=(2hati+3hatj)N and vecF_(2)(-2hatk+3hatj)N.` If the body is displaced from `A(3m, -2m, 1m)` to B `(-1m, +2m,-3m),` then work done on the body isA. 40 JB. `-40 J`C. 24 JD. `-24J` |
| Answer» Correct Answer - C | |
| 1176. |
An object is displaced from point `A(2m,3m,4m)` to a point `B(1 m,2 m,3 m)` under a constant force ` F=(2hati + 3hatj + 4hatk)N`. Find the work done by this force in this process. |
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Answer» `W=int_(r_i)^(r_f) F*dr=int_((2m, 3m, 4,))^((1m, 2m, 3m,))(2hati + 3hatk + 4hatk)*(dxhati + dyhatj + dzhatk)` `=[2x + 3y + 4z] _((2m, 3m, 4m)) ^((1m, 2m, 3m,))=-9J` Alrernate Solution Since, `F = "constant"`, we can also use. `W=F*S` Here, `S=r_f-r_i=(hati+2hatj+3hatj+3hatk)-(2hati+3hatj+4hatk)` `=(-hati-hatj-hatk)` `W=(2hati+3hatj+4hatk)*(-hati-hatj-hatk)`ltbrge-2-3-4=-9J`. |
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| 1177. |
A constant force `F=(hati+3hatj+4hatk)N` acts on a particle and displace it from (-1m,2m,1m)to(2m,-3m,1m). |
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Answer» Given, `F=hat i+3hatj+4hatk` and Initail position of the particle is given by `s_(1)=-hat i+2hatj+hat k` Final position, `s_(2)= 2hati-3hatj+hatk` Displacement of the particle , `s=s_(2)-s_(1)=(2hati-3hatj+hat k)-(hat i+2hatj+hat k)` ltbrlt `=3hati-5hatj+0hatk=3hatj-5hatj` Work done by the force F `W=F.s=(hat i+3hatj+4hatj).(3hati-5hatj)=(3-15)=-12 J` |
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| 1178. |
Water falls from a height of `60 m` at the rate `15 kg//s` to operate a turbine. The losses due to frictional forces are `10%` of energy . How much power is generated to by the turbine? (g=10 m//s^(2))`.A. 12.3 kWB. 7 kWC. 8.1 kWD. 10.2 kW |
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Answer» Correct Answer - C (c ) Power given to turbine `=(mgh)/(t)` `P_("in")=((m)/(t))xxgxxh rArrP_("in")=15xx10xx60` `rArrP_("in")=9000" W" rArr P_("in")=9" k W"` As effeciency of turbine is 90% therefore power generated =90 % of 9`" k W"` `P_(out)=9xx(90)/(100)rArr P_(out)=8.1" kW"`. |
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| 1179. |
An engine is attahed to a wagon through a shock absorber of length 1.5m. The system with a total mass of 50,000kg is moving with a speed of `36kmh^(-1)` when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by `1.0m`. If `90%` of energy of the wagon is lost due to friction, calculate the spring constant. |
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Answer» Here,`l=1.5m, m=50000kg=5xx10^(4)kg` `upsilon=36kmh^(-1)=(36xx1000)/(60xx60)ms^(-1)=10ms^(-1), x=1.0m,k=?` Total KE`=(1)/(2)m upsilon^(2)=(1)/(2)xx5xx10^(4)(10)^(2)=2.5xx10^(6)J` As `90%` of energy of wagon is lost due to friction, therefore, energy transferred to shock absorber, `E=(10)/(100)xx2.5xx10^(6)=2.5xx10^(5)J` From `E=(1)/(2)kx^(2)` `k=(2E)/(x^(2))=(2xx2.5xx10^(5))/(1^(2))=5xx10^(5)N//m` |
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| 1180. |
The turbine pits at Niagra Falls are 50m deep. The average horse power develop is 5000. If the efficiency of the generator `85%` , how much water passes through the turbine per minute ? Take `g=10m//s^(2)`. |
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Answer» Here, `h=50m, H.P.=5000, eta=85%`, `m_(2)` useful power `=5000HP=5000xx746 wat t`. Total power generated `=(5000xx746)/(85%)=(500xx746xx100)/(85)` `=4.388xx10^(6) wat t` Total work done by the falling water in 1 minute `mgxxh=4.388xx10^(6)xx60J` `m=(4.388xx10^(6)xx60)/(gh)=(4.388xx10^(6)xx60)/(10xx50)` `m=5.39xx10^(5)kg` |
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| 1181. |
A railway carriage of mass 9000kg moving with a speed of `36kmh^(-1)` collides with a stationary carriage of the same mass. After the collision, the two get coupled and move together. What is this common speed ? What type of collision is this? |
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Answer» Here, `m_(1)=9000kg.` `u_(1)=36km//h=10m//s` `m_(2)=9000kg, u_(2)=0, v_(1)=v_(2)=v=?` From conservation of linear momentum, `(m_(1)+m_(2))v=m_(1)u_(1)+m_(2)u_(2)=m_(1)u_(1)` `v=(m_(1)u_(1))/((m_(1)+m_(2)))=(9000xx10)/(9000+9000)=5ms^(-1)` Total KE before collision `=(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)=(1)/(2)xx9000xx10^(2)` `=4.5xx10^(5)J` Total KE after collision `=(1)/(2)(m_(1)+m_(2))v^(2)` `=(1)/(2)(9000+9000)xx5^(2)=2.25xx10^(5)J` As KE after collision `lt` KE before collision `:. ` Collision is inelastic. |
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| 1182. |
A rain drop of radius 2mm, falls from a height of 500 m above the ground. It falls with decreasing acceleration due to viscous resistance of air until half its original height. It attains its maximum (terminal ) speed, and moves with uniform speed there after. What is the work done by the gravitational force on the drop in the first half and second half of its journey ? Take density of water `=10^(3)kg//m^(3)`. What is the work done by the resistive force in the entire journey if its speed on reaching the ground is `10ms^(-1)` ? |
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Answer» Here, r=2 mm `=2 xx 10^(-3)m` Distance moved in each half ot the journey, S=`500//2` = 250 m Density of water , p=`10^(3)kg//m^(3)` Mass of rain drop = volume of drop x density `m=4//3pir^(2)xx p= 4//3 xx 22//7(2 xx 10^(-3))^(3) xx 10^(3)= 3.35 xx 10^(-5)kg` `therefore W= mg xx s = 3.35 xx 10^(-5) xx 9.8 xx 250= 0.082` J Note, whether the drop moves with decreasing acceleration or with uniform speed, work done by the gravitational force on the drop on reaching the ground. `E_(1) = mgh= 3.35 xx 10^(-5) xx 9.8 xx 500 = 0.164`J Actual energy, `E_(2) = 1//2mv^(2)= 1//2 xx 3.35 xx 10^(-5)(10^(2)= 1.675 xx 10^(-3)J` Work done by the resistive forces, `W=E_(1)-E_(2) = 0.164 - 1.675 xx 10^(-3)` W = 0.1623 Joule. |
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| 1183. |
Rain derops each of mass m falling from rest in air experience an upward force given by f= - bv where b is a constant and v is velocity of the drop. Using work energy theorem, derive an expression for (A) `v=f(t)` i.e. velocity function of time. (B) `(dE)/(dt)=f(t)` where E = Total machanical energy of the rain drop ( C) Draw the graphs `(dE)/(dt.(dU)/(dt) and P_("viscous")` as a function of time U is potential energy and P represent power ? |
| Answer» `(A)v=(mg)/(b)(1-e^((-bt)/(m))),(b)(dE)/(dt)=(m^(2)g^(2))/(b)(1+e^((-bt)/(m)))` | |
| 1184. |
Is it practically possible to have situtations where `(E-V)lt0`? |
| Answer» No, because `(E-V)` represents K.E. which cannot be negative. | |
| 1185. |
What is the source of KE of falling rain drops ? |
| Answer» Potential energy of falling rain drops goes on decreasing and their KE goes on increasing. | |
| 1186. |
What is work done in holding a 15kg suitcase while waiting for a bus for 15 minutes ? |
| Answer» Work done is zero, because displacement is zero. | |
| 1187. |
The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in figure. If the length of the pendulum is 1m, calculate (a) the height to which bob `A` will rise after collision. (b) the speed with which bob `B` starts moving. Neglect the size of the bobs and assume the collision to be elastic. |
| Answer» (a) As the collision is elastic and two balls have same mass, therefore, ball `A` transfers its entire linear momentum to `B` and does not rise at all. | |
| 1188. |
The bob A of a simple pendulum released from `30^(@)` to the vertical hits another bobo B of the same mass at rest on a table as shown in figure. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic. |
| Answer» The bob A shall not rise. This is because when two bodies of same mass undergo an elastic collision, their velocities are interchanged. After collision, ball A will come to rest and the ball B would move with the velocity of A. figure,. Thus the bob A will not rise after the collision | |
| 1189. |
A pump on the ground floor of a building can pump of water to fill a tank of volume `30 ms^(3)` in `15 min`. If the tank is `40m` above the ground and the efficiency of the pump is `30 %` , how much electric power is consumed by the pump? `("Take" g = 10 ms^(2))` |
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Answer» Here, volume of water = `30ms^(3)`, t=15min = `15 xx 60=90s` h=40m, n=`30%` As the density of water = p = `10^(3) kg m^(-3)` Mass of water pumped, m = volume xx density = `30 xx 10^(3)` kg Actual power consumed or output `p_(0) = W//t = mgh//t` `rArr p_(0)(30 xx 10^(3) xx 9.8 xx 40)//900 = 13070 watt` If pi is input power (required), then as `eta = p_(0)//p_(i)rArr p_(i)=p_(0)//eta` = `13070(30//100) = 43567 W = 43.56` KW |
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| 1190. |
The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic. |
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Answer» Bob A will not rise at all In an elastic collision between two equal masses in which one is stationary, while the other is moving with some velocity, the stationary mass acquires the same velocity, while the moving mass immediately comes to rest after collision. In this case, a complete transfer of momentum takes place from the moving mass to the stationary mass. Hence, bob A of mass m, after colliding with bob B of equal mass, will come to rest, while bob B will move with the velocity of bob A at the instant of collision. |
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| 1191. |
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head on by another ball bearing of the same mass moving initially with a speed v, figure,. If the collision is elastic, which of the following is a possible result after collisioin? |
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Answer» Let m be the mass of each ball bearing. Before collision, total K.E. of the system is `=(1)/(2)mv^(2)+0=(1)/(2)mv^(2)` After collision. K.E. of the system is Case I, `E_(1)=(1)/(2)(2m)(v//2)^(2)=(1)/(4)mv^(2)` Case II, `E_(2)=(1)/(2)mv^(2)` Case III, `E_(3)=(1)/(2)(3m)(v//3)^(2)=(1)/(6)mv^(2)` We observe that K.E. is conserved only in case II. Hence case II is the only possibility. |
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| 1192. |
Which physical quantity is conserved in both, elastic and inelastic collision? |
| Answer» Linear momentum is conserved in both elastic and inelastic collisions. | |
| 1193. |
In which, elastic or inelastic collision, the momentum is conserved ? What about K.E.? |
| Answer» Momentum is conserved in both the types of collisions. But K.E. is conserved only in elastic collisions. | |
| 1194. |
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head on by another ball bearing of the same mass moving initially with a speed v, figure,. If the collision is elastic, which of the following is a possible result after collisioin? |
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Answer» Let m be the mass of each ball bearing. Before collision, total K.E of the sytem `1/2mv^92)+0=1//2mv^(2)` After collision, K.E of the system is Case I, `E_(1)=1//2(2m)(v//2)^(2)= 1//4mv^(2)` Case II, `E_(2) = 1//2mv^(2)` Case III, `E_(3) = 1//2(3m)(v//3)^(2)=1//6mv^(2)` Thus, case II is the only possibility since K.E is conserved in this case. |
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| 1195. |
Two identical ball bearings in contact with each other and resting on a frictionless table are hit heat-on by another ball bearing of the same mass moving initially with a speed `V` as shown in figure. If the collision is elastic, which of the following (figure) is a possible result after collision ?A. B. C. D. |
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Answer» (b) When two bodies of equal masses colides elastically their velocities are interchanges. When ball 1 collides with ball-2 then velocity of ball-1. `v_(1)` becomes zero and velocity of ball-2, `v_(2)` becomes v,i.e. similarity When ball 2 collides will ball 3 `{:(,v_(1)=0,Rightarrow v_(2)=v),(,v_(2)=0,v_(3)=v):}` |
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| 1196. |
Statement-1 : Kinetic energy is conserved in both, perfectly elastic and perfectly inelastic collisions. Statement-2 : Because linear momentum is conserved in both.A. Statement - 1 is ture , statement -2 is true, and statement-2 is correct explanation of statement -1.B. Statement-1 if ture, statement-2 is true, but statement -2 is not a correct explanation of statement-1.C. Statement-1 is true, but statement -2 is false.D. Statement-1 is false, but statement-2 is true. |
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Answer» Correct Answer - D Kinetic energy is not conserved in a perfectly inelastic collision, as some of the total energy amy be converted into heat energy etc. Statement`-1` is false, but statement `-2` is true. |
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| 1197. |
A shell is fired from a cannon with a velocity `v (m//sec.)` at an angle `theta` with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (in `m//sec.`) of the other piece immediately after the explosion isA. `3vcostheta`B. `2vcostheta`C. `(3)/(2)vcostheta`D. `(sqrt(3))/(2)vcostheta` |
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Answer» Correct Answer - A |
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| 1198. |
A `2kg` block slides on a horizontal floor with the a speed of `4m//s` it strikes a uncompressed spring , and compresses it till the block is motionless . The kinetic friction force is compresses is `15 N` and spring constant is `10000 N//m` . The spring byA. 5.5 cmB. 2.5 cmC. 11.0 cmD. 8.5 cm |
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Answer» Correct Answer - C `(1)/(2)mv^2=(1)/(2)kx^2+W_f` `(1)/(2)(2)(4)^2=(1)/(2)(10000)x^2+f_(k)x` `16=5xx10^3x^3+15x` `5xx10^3x^3+15x-16=0` `x=(-15+-sqrt((15)^2+4xx5xx10^2xx16))/(5xx10^3)` `=(11)/(100)m=11cm` |
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| 1199. |
The displacement of a body of mass `2 kg` varies with time `t` as `S = t^(2) + 2t`, where `S` is in seconds. The work done by all the forces acting on the body during the time interval `t = 2s` to `t = 4s` isA. 36 JB. 64 JC. 100 JD. 120 J |
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Answer» Correct Answer - B |
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| 1200. |
A block of mass m is hung vertically from an elastic thread of force constant `mg//a`. Initially the thread was at its natural length and the block is allowed to fall freely. Kinetic energy of the block when it passes through the equilibrium position will beA. mgaB. `(mga)/(2)`C. zeroD. 2mga |
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Answer» Correct Answer - B |
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