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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
What is the torque of a force 3hati+7hatj+4hatk about the origin, if the force acts on a particle whose position vector is |
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Answer» `hati - 5hatj + 8 hatk` Torque , `vectau = vecr xx vecF` `r = |{:(hati , hatj , hatk) , (2 , 2 , 1) , (3, 7 , 4):}| = hati(8-7) - hatj (8-3) + hatk (14 - 6) = hati - 5hatj + 8hatk` |
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| 2. |
The total number of air molecules (inclusive of oxygen ,nitrogen ,water vapour and other contituents)in a room of capacity 20 m^(3) at a temerature of 27^(@)C and 1 atm pressure. |
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Answer» Solution :Givne data VOLUME `(V)=20m^(3)` Temperature `(T)=27+273=300K` Pressure (P) =1 atm `=1.01 XX 10^(5)PA` TOTAL NUMBER of air molecules `N=(PV)/(KT)` `=(1.02xx 10^(5) xx 20)/(1.38 xx 10^(-23) xx 300)=4.87 xx 10^(26)` |
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| 3. |
A body of mass 2kg is moving along positive X - direction with a velocity of 5 ms^(-1) Now a force of 10sqrt(2) NN is applied at an angle 45° with X - axis. Its velocity after 3s is, |
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Answer» ` 20 ms^(-1 )` |
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| 4. |
Masses of atomic// subatomic particles are measured using a …………….. . |
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Answer» |
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| 5. |
How does the kinetic energy of a body change if its momentum is doubled? |
| Answer» SOLUTION :`K = P^2/(2m)`. As `K prop P^2`, its KINETIC ENERGY becomes FOUR times the initial value. | |
| 6. |
{:(,"List - I",,"List - II"),((a),"Centrifugal force",(e ),"Along the axis of rotation"),((b),"Centripetal force",(f),"Towards the centre of rotation"),((c ),"Tangential force",(g),"Away from the centre of rotation"),((d),"Angular velocity",(h),"Changes the angular velocity"):} |
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Answer» a-h, B-g, c-f, d-e |
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| 7. |
If force |F|, velocity |v| and time |T| are taken as to fundamental units then the dimensions of mass are |
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Answer» `FV^(-1) T` `m = (F.dt)/(dv)` Dimensionally the mass is `FV^(-1) T` |
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| 8. |
A force of 2i + 3j + 2k N acts a body for 4s and produces a displacement of 3i + 4j + 5k m calculate the power? |
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Answer» 5W |
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| 9. |
Does the escape velocity depend on the location from where it is projected? |
| Answer» SOLUTION :Yes, It DEPENDS on the LOCATION from where satellite is PROJECTED. | |
| 10. |
A stone of mass 0.05 kg is thrown in vertical by upward direction (take g = 10 mcdot s^(-2)). Neglect air friction. The net force acting on the stone during its downwards motion is |
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Answer» 0.5 N, upward |
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| 11. |
A listener is at rest with respect to the souce of sound. A wind starts blowing along the line joining the source and the observer. Which of the following quantities do not change? |
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Answer» Frequency |
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| 12. |
The length of an iron rod is found to be smaller on a hot summer day than on a cold winter day as measuredby the same aluminium scale. Is it correct to conclude that iron shrinks on heating? (alpha of iron =12xx10^(-6)//""^(0)C and alpha of aluminium =24xx10^(-6)//""^(0)C). |
| Answer» Solution :No, because during WINTER IRON CONTRACTS LESS than aluminium. So, the iron rod appears LONGER during winter. | |
| 13. |
A rod of weight W is supported by two parallel knife edges A and B is in equilibrium in a horizontal position. The knives are at a dist once d from each other. The center of mass of the rod is at distance x from A. The normal reaction on A is |
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Answer» `(W(d-X))/(d)` |
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| 14. |
Match the column I with column II |
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Answer» A-p,B-q,C-r,D-s `=("Charge"xx"charge")/(4pixx"electrical FORCE"xx"distance"^(2))` `[epsi_(0)]=([AT][AT])/([MLT^(-2)][L]^(2))=[M^(-1)L^(-3)T^(4)A^(2)]` `A-s` Radiant flux `=("energy emitted")/("Time")=([ML^(2)T^(-2)])/([T])=[ML^(2)T^(-3)]` Resistivity`=("RESISTANCE"xx"area")/("Length")` `[rho]=([ML^(2)T^(-3)A^(-2)][L^(2)])/([L])=[ML^(3)T^(-3)A^(-2)]` `C-q` Hubble constant `=("Recession speed")/("Distance")=([LT^(-1)])/([L])=[ML^(0)L^(0)T^(-1)]` |
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| 15. |
A three storey building of height 100m is located on Earth and a similar building is also located on Moon. If two people jump from the top of these buildings on Earth and Moon simultaneously, when will they reach the ground and at what speed?(g=10ms^(-2)) |
Answer» Solution : For both persons, the KINEMATIC equations are the same, with `u=0,a_(c)=g` and `a_("moon")=(g)/(6)`. Then `a_(E)=g` and `a_(m)=(g)/(6)` For a person on Earth, `V_("earth")=sqrt(2gh)=sqrt(2xx10xx100)` HENCE, `V_("earth")=sqrt2000ms^(-1)` givens the velocity at the GROUND, on Earth. Similarly, for a person on the Moon, `V_("earth")=sqrt((2gh)/(6))=(sqrt2000)/(sqrt6)MS^(-1)` The person on Earth reaches ground with greater velocity than the person on the. |
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| 16. |
For angle of projection of a projectile at angles and (45^(@)+theta), the horizontal range described by the projectile are in the ratio of |
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Answer» `2:1` For angle ofprojection `(45^(@)-theta)`.the horizontal range is `R_(1)=(u^(2)sin [2(45^(@)-theta)])/(g)=(u^(2)sin(90^(@)-2theta))/(g)=(u^(2)cos 2 theta)/(g)` `:.(R_(1))/(R_(2))=(u^(2)cos 2 theta//g)/(u^(2)cos 2 theta//g)=(1)/(1)` |
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| 17. |
One end of a uniform rod of mass m and length / is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity omega. The force exerted by the clamp on the rod has a horizontal component. |
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Answer» `m OMEGA^(2)L` |
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| 18. |
A soap bubble is blown to a radius of 3 cm. If it is to be further blown to a radius of 4 cm what is the work done ? (Surface tension of soap solution = 3.06xx10^(-2)Nm^(-1)) |
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Answer» Solution :Initial radius of the soap bubble `R_(1)=3cm=3xx10^(-2)m`. Final radius of the soap bubble `R_(2)=4cm=4xx10^(-2)m` work done in blowing a soap bubble from radius `R_(1)` to `R_(2)=W=8pi(R_(2)^(2)-R_(1)^(2))S` `=8XX(22)/(7)xx3.06xx10^(-2)(16-9)xx10^(-4)` `=176xx3.06xx10^(-6)J=539.6xx10^(-6)J` |
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| 19. |
What is Degree of freeedom ? |
| Answer» SOLUTION :The degree of freedomfor a dynamic system is THENUMBER of directions in whichit canmovefreelyor the numberof COORDINATES requiredto describecompletelythe POSITON and CONFIGURATIONOF the system. | |
| 20. |
The radius of a metal sphere at room temperature T is R , and the coefficient of linear expansion of the metal isalpha. The sphere is heated a little by a temperature Delta T so that its new temperature is T + Delta T. The increase in the volume of the sphere isapproximately |
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Answer» `2pi R ALPHA Delta T` |
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| 21. |
Explain and draw the graphs of kinetic energy, potential energy and total energy as a function of time. |
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Answer» SOLUTION :Kinetic energy of SHM particle `K(t) =(1)/(2) kA^(2) sin^(2) (omega t+phi)` Potential energy `U(t) = kA^(2) cos^(2) (omega t+phi)` Mechanical energy `E= K(t) +U(t)` `therefore (1)/(2) kA^(2) [sin^(2) (omega t+ phi)+cos^(2)(omega t +phi)]` `therefore E= (1)/(2)kA^(2)`. Here, in table values of kinetic, potential and total energy in terms of period of SHM particle is shown and by using these values a graph of a energy versus time is shown.   Following points are clear from this graph : (1) In SHM both kinetic and potential energy are positive HENCE total energy is also positive. (2) In SHM, kinetic energy becomes two time maximum in every period and becomes two times zero. This maximum kinetic energy is equal to the total energy. (3) In SHM, potential energy becomes two times maximum in every period and BECOME two time zero. This maximum potential energy is eaual to the total energy. (4) The increase or decrease in kinetic energy from energy time of `(T)/(4)` in SHM is equal to decrease or increase in potential respectively. (5) At any time the sum of kinetic and potential energy is constant and it is mechanical energy. Hence mechanical energy does not depend on time. |
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| 22. |
A springs (spring constant k) is cut into 4 equal parts and two parts are conneced in parallel.What is the effective spring constant? |
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Answer» 4k |
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| 23. |
Can we find the correct temperature of a body on Celsius using a faulty thermometer ? |
| Answer» Solution :YES, if we KNOW the FIXED points of the faulty THERMOMETER. | |
| 24. |
A balloon with mass m is descending down with an acceleration a where a |
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Answer» `(2MA)/(g+a)` ` thereforemg -F =ma .......(i)` Letmass ` m_0`beremovedfromthe balloonso THATIT startsmovingupwithan accelerationa. then ` F- (m-m_0) g= ( m-m_0)a` `f-mg + m_0 g = ma- m_0a `....(ii) addingeqn. (i )andeqn(ii)we get ` m_0g - 2ma- m_0,m_0 g + m_0 a= 2ma,m_0 ( g+a ) = 2ma ` ` m_0= (2ma )/(g+a)` 
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| 25. |
State the number of significant figures in the following (i) 600800 (ii) 400 (iii) 0.007 (iv) 5213.0 (v) 2.65 xx 10^(24)m (vi) 0.0006032 |
| Answer» Solution :(i) four (ii) one (III) one (IV) five (v) THREE (VI) four | |
| 26. |
In the above problemrange of body is m. [g=10ms^(-2)] |
| Answer» Answer :A | |
| 27. |
A stationary bomb of mass 10 kg suddenly explode in two fragment of 4 kg and 6 kg. Ratio of their velocity will b e ..... |
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Answer» `3:2` massof secondfragment`m_(2) = 6` kgand velocityBy lawof conservation of linearmomentum `MV = m_(1) v_(1) +m_(2) v_(2)` `M xx 0 = m_(1) v_(1) + m_(2) v_(2)` `m_(1) v_(1) -m_(2)v_(2)` `(v_(1))/(v_(2)) =-(m_(2))/( m_(1))` `|(v_(1))/(v_(2))| = (6)/(4)` `v_(1) :v_(2)= 3: 2` |
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| 28. |
Particle P and Q of mass 20 g and 40 g respectively are simultaeously projected from points A and B on the ground. The initial velocities of Pand Q make 45^(@) and 135^(@) angles respectively with the horizontal AB as shown in the fig. Each particle has an initial speed of 49m/s. The separation AB is 245 m. Both particles travel in the same verical plane and undergo a collision. After collision P retraces its path. Determine the position of Q when it hits ground. How much time after the collision does the particle Q take to reach the ground? (Take g=9.8m//s^(2)) |
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Answer» |
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| 29. |
A stone dropped from the top of a tower of height 300m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340ms^(-1) ? (g= 9.8ms^(-2)) |
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Answer» |
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| 30. |
If vecA xx vecB = vecC+ vecD, them select the correct alternative: |
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Answer» `vecB` is parallel to `vecC + VECD.` i.e., `vec(A).(vec(C)+vec(D))=0orvec(A). vec(C)+vec(A).vec(D)=0` or A (component of `vec(C)` along `vec(A)`) + A(component of `VCE(D)` along `vec(A)`)=0 or component of `vec(C)` along `vec(A)` =- component of `vec(D)` along `vec(A)` |
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| 31. |
A thin prism having refracting angle 10^@ is made glass of refractive index 1.42. This prism is combined with another thin prism of glass of refractive index 1.7. This combination produces dispersion without deviation. The refracting angle of second prism should be: |
| Answer» ANSWER :C | |
| 32. |
Coefficient of friction between the block and the surface in each of the given figures being 0.4 , match Column - I and II . (Take g =10 m//s ^(2)) |
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Answer» `tan theta = (ma)/(MG) = (a)/(g) , "" therefore theta = tan^(-1) ((a)/(g))` 
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| 33. |
If m is the mass and E the K.E of a body then its linear momenta is |
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Answer» `msqrtE` |
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| 34. |
A wheel ( to be considered as a ring ) of mass m and radius R rolls without sliding on a horizontal surface with constant velocity v. It encounters a step of height R/2 at which it ascends without sliding |
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Answer» the angular velocity of the ring just after it comes in contact with step is 3v/4R |
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| 35. |
Statement A : Action and reaction act on two different bodiesStatement B : Action, reaction never cancel each other |
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Answer» A & B are CORRECT |
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| 36. |
A man runs along a horizontal road on a rainy day. His umbrella vertical in order to protect himself from the rain then rain actually |
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Answer» COMING from front of a man |
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| 37. |
Give the relation aroone |
| Answer» Solution :When thermal expansion is not allowed in a solid, the corresponding STRESS DEVELOPED in it is known as thermal stress. The solid acquires a compressive strain due to the external FORCES, provided by (say) the rigid support at one END of it. | |
| 38. |
On a rainy day skidding takes place along a curved path. Why? |
| Answer» Solution : As the friction between the TYRES and road REDUCES on a RAINY DAY. | |
| 39. |
A piece of wood floats in water with two-thirds of its volume submerged in water. In a certain oil it has 0.95 of its volume submerged. What is the density of wood and the oll? |
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Answer» Solution :(i) Wood FLOATING in WATER Volume of wood `XX`density of wood Volume of immersed PORTION of the wood `xx` density of water ` "" V xx D = (2)/(3)V xx 1000 ,D = 666. 67 kg //m^(3)` ` "" D = 666. 67 kg //m^(3) ` (ii)Wood floating in oil ` "" V xx D=v xx rho , V xx ( 2000)/(3)= 0. 95 V xx rho` Density of oil ` = ( 2000)/( 3xx0.95)701.7 kg//m^(3)` |
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| 40. |
The potential energy of projectile of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is 20 ms^(-1), its time of flight is (g = 10 ms^(-2)) |
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Answer» 2S |
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| 41. |
A body is projected at an angle 30^(@) to the horizontal with speed of 30 ms^(-1). The angle made by the velocity vector with the horizontal after 1.5 s is(g = 10 ms^(-2)) |
| Answer» ANSWER :A | |
| 42. |
A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R. A smooth pulley of small radius is fastended to the table. Two masses m and 2m placed on the table are connected through a string over the pulley. Initially the masses are held by a person with the string along the outward radius and then the system is released from the (with respect to the cabin). Find the tension in the string. |
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Answer» `(4)/(3)momega^(2)R` |
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| 43. |
A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal end containing air at the same pressure P. When the tube is held at an angle of 60^@ with the vetical direction, the length of the air column above and below the mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure P in centimeters of mercury. (The temperature of the system is kept at 30^@C). |
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Answer» <P>`75.4` When the rube is held at `60^(@)` with the VERTICAL, the lengths of air columns at the bottom and the top are `44.5` cm and 46 cm respectively. If `P_(1) and P_(2)` are their pressures. then `P_(1)-P_(2)5 cos60^(@)=5xx1/2=5/2` cm of Hg Using Boyle's LAW for constant temperature, `PV =P_(1)V_(1)=P_(2)V_(2)` `PxxAxx45.25=P_(1)xxAxx44.5=P_(2)xxA46` `therefore(Pxx45.25)/(44.5)-(Pxx45.25)/(46)=5/2` `or P(5xx44.5xx46)/(2xx45.25xx1.5)=75.4` cm of Hg |
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| 44. |
In a cyclic process, the internal energy of the gas |
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Answer» increases |
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| 45. |
What does the universal gas constant R signify? Give its value. |
| Answer» SOLUTION :The universal GAS constant R signifies the wordone by (or on) a gas PER mole per kelvin. Its value is `R=8.314" J mol"^(-1) K^(-1)`. | |
| 46. |
A wad of sticky clay of mass m and velocity v_(i) is fired at a solid cylider of mass M and radius R figure. The cylinder is initially at rest and is mounted on a fixed horizontal axle that runs through the centre of mass. The line of motion of the projectile is perpendicular to the axle and at a distance d, less thant R, from the centre |
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Answer» Angular velocity just after COLLISION is `OMEGA=(2mv_(i)d)/((M+2m)R^(2))` |
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| 47. |
Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be (MR^(2))/(4). Find its moment of inertia about an axis normal to the disc and passing through a point on its edge. |
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Answer» Solution :Given `I_(x) =(MR^(2))/(4)and I_(y)=(MR^(2))/(4)` According to PERPENDICULAR -AXES theorem, `I_(z)=I_(x)+I_(y)=2.(MR^(2))/(4)` `=(1)/(2)MR^(2)` According to parallel -axes theorem THEMOMENT of inertia of the disc about the axis AB, normal to the disc and passing through a point on its edge, `I-I_(cm)+MR^(2)=I_(z)+MR^(2)=(1)/(2)MR^(2)+MR^(2)=(3)/(4)MR^(2)` 
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| 48. |
(A) By roughening the surface of a sheet its transparency can be reduced. (R ) Glass sheet with rough surface reflects more light. |
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Answer» Both A and R are true and R is the CORRECT explanation of A |
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| 49. |
294 joules of heat are required to rise thetemperature of 2 mole of an ideal gas at constant pressure from 30^(@)C to 35^(@)C. The amount of heat required to rise the temperature of the same gas through the same range of temperature at constant volume (R = 8.3 Joules/mole/K) is |
| Answer» ANSWER :B | |