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.
| 39951. |
A heavy iron rod of weight 'W' is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle 'theta' with the horizontal. What is the weight experienced by the man ? |
|
Answer» W `sintheta` |
|
| 39952. |
The acceleration due to gravity at the poles is 10ms^(-2) and equitorial radius is 6400km for the earth. Then the angular velocity of rotation of the earth about its axis so that the weight of a body at the equator reduces to 75% is |
|
Answer» `1/1600 "RADS"^(-1)` |
|
| 39953. |
The situations are shown in fig(a) & (b), in each case, m_(1)=3kg and m_(2)=4kg. If a_(1), a_(2) are the respective accelerations of the blocks in these situations, then the values of a_(1) and a_(2) are respectively [g=10 ms^(-2)] |
|
Answer» `(10)/(7)MS^(-2), (25)/(7)ms^(-2)` |
|
| 39954. |
If a cylinder of diameter 1.0cm at 30°c is to be slide into a hole of diameter 0.9997 cm in a steel plate at the same temperature, then minimum required rise in the temperature of the plate is [coefficient of linear expansion of steel = 12 xx 10^(-6) //^(@) C ] |
|
Answer» `25^(@) C` |
|
| 39955. |
What will be the difference in temperatures between two sides of a metallic plate 10 mm thick if heat is conducted at the rate of 5 xx 10^5cal/min/m^2 . K for metal is 0.7 cal s^(-1)cm^(-1) ""^@C^(-1) . |
|
Answer» SOLUTION :`Delta x = 10 MM = 1 cm` `K = 0.7 CAL s^(-1) cm^(-1) K^(-1)` ` Q = 5 xx 10^5 cal , A = 1m^2 = 10^4cm^2` t = 1 min = 60 s ` Q = (KA(T_1 - T_2)t)/(Delta x)` ` rArr T_1 - T_2 = (Q Delta x)/(KA t) = (5 xx 10^5 xx 1)/(0.7 xx 10^4 xx 60)` ` T_1 - T_2 = 1.19^@C` |
|
| 39956. |
Consider the motion of the tip of the minute hand of a clock . In one hour (i) the displacement is zero (ii) the distance covered is zero (iii) the average speed is zero (iv) the average velocity is zero |
|
Answer» (i), (ii) are CORRECT velocity `=("displacement")/("time") = 0` |
|
| 39957. |
A uniform cylinder of height h and radius r is placed with its circular face on a rough inclined plane and the inclination of the plane to the horizontal gradually in creased. If mu is the coefficeint of friction, then under what conditions the cylinder will a) slide before topping b) topple before sliding |
Answer» Solution : (a) The Cylinder will slide if `Mgsintheta gt mu Mgcosthetaimpliestantheta gt mu`……….`(1)` The cylinder will TOPPLE if `(Mgsintheta)(h)/(2) gt (Mg costheta)rimpliestantheta gt (2r)/(h)`…………`(2)` THUS, the condition of sliding is `tantheta gt mu` and condition of toppling `tantheta gt (2r)/(h)`. Hence, the cylinder will slide before toppling if `mu LT (2r)/(h)` (b) The cylinder will topple before sliding if `my gt (2r)/(h)` |
|
| 39958. |
Three equal masses .m. are placed at the three vertices of an equilateral triangle of side a. Find the gravitational force exerted by this system on another particle of mass ‘m. placed a) at the mid point of a side and (b) at the centroid of the triangle. |
| Answer» SOLUTION :`a) 4Gm^2 // 3a^2 , B) 0` | |
| 39959. |
A piece of brass (Cu and Zn) weighs 12.9 g in air. When completely immersed in water, it weights 11.3 g. The relative densities of Cu and Zn are 8.9 and 7.1 respectivey. Calculate the mass of copper in the alloy |
|
Answer» 6.6 gm |
|
| 39960. |
When a solid sphere is undergoing pure rolling, the ratio of translational kinetic energy to rotational kinetic energy is |
|
Answer» `2:5` |
|
| 39961. |
Give the vector form of the relation between v and omega. |
| Answer» SOLUTION :The vector RELATION between linear velocity and ANGULAR velocity `omega` is `barv=vecrxxbaromega` | |
| 39962. |
The resultant of two forces, one double the other in magnitude is perpendicular to the smaller of the two forces. The angle between the two forces is |
|
Answer» `150^(@)` |
|
| 39963. |
Two satellites S_(1) and S_(2) are revolving around a planet P in circular orbits of radii 4R and 9R respectively. If the speed of satellite S_(1) is 2V, then calculate the speed of satellite S_(2). |
|
Answer» Solution :As we know orbital speed `V_(0) = SQRT((GM)/(R ))` For `S_(1), ""2V = sqrt((GM)/(4R))` …(i) For `S_(2), ""V. = sqrt((GM)/(9R))` …(ii) Dividing (i) by (ii) we get `(V.)/(2V) = sqrt((GM)/(9R)).sqrt((4R)/(GM))` `v. = 2V.2//3 = 4V//3` |
|
| 39964. |
A particle executes SHM represented by the equation , y= 0.02sin(3.14t+ (pi)/(2)) meter. Find frequency |
|
Answer» SOLUTION :COMPARING EQUATION `y= 0.02sin(3.14t+ (pi)/(2))`with the general from of the equation , `y= A sin(omegat+phi_(0))` Frequency `v= (1)/(T)= (1)/(2)Hz= 0.5 Hz` |
|
| 39965. |
In the previous problem, If 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? |
| Answer» Solution :MERCURY will RISE in the arm containing spirit, the difference in LEVELS of mercury will be 0.221 CM. | |
| 39966. |
An object of mass 0.20kg executes simple harmonic oscillations along X-axis with a frequency of (25//pi) Hz. At the position x=0,04 the object has kinetic energy of 0.5J and potential energy 0.4J. The amplitude of oscillations is: |
|
Answer» 0.06 m |
|
| 39967. |
(A) : For a satellite revolving very near to earth's surface the time period of revolution is given by 1 hour 24 minutes. (R) : The period of revolution of satellite depends upon its height above the earth's surface. |
|
Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
|
| 39968. |
A mass of 2kg is suspendedfrom a fixed point by a wire of length 3m and diameter 0.5 mm. Initially the wire is just unstretched, the mass resting on a fixed support. By hoH' much must the temperature fall if the mass is to be entirely supported by the wire? Given that Y of the wire = 206 G Pa, alpha = 11 xx 10^(-6)//^(0) C . |
|
Answer» Solution :When the temperature DECREASES the wire contracts in length and LIFTS the mass up. But the weight begins to produce a stress and counteracts the contraction PRODUCED. Hence the contraction due to cooling is equal to the stretching produced by the weight W. `therefore DELTA L = (WL)/(AY) = (2 xx 9.8 xx 3)/(pi (0.25)^(2) xx 10^(-6) xx 206 xx 10^(9))` Now the contraction due to cooling ` = L alpha d theta = 3 xx 11 xx 10^(-6) xx d theta` Since the contraction due to cooling = extension produced by the weight. On solving `d theta = 44^(@)`C |
|
| 39969. |
Let I_(A) and I_(B) be moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not. |
|
Answer» `I_(A)LT I_(B)` |
|
| 39970. |
can a physical quantity having both magnitude and direction be a vector? |
| Answer» Solution :No, it is not necessary that a physical quantity having both MAGNITUDE and direction is a vector . For example, CURRENT FLOWING in a conductor is having both magnitudeand direction but is a scalar quantity because the laws of vector ADDITION but is a scalar quantity because the laws of vecrtor addition are not applicable to ELECTRIC current . | |
| 39971. |
Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is. . . . . .. . |
|
Answer» `sqrt(2K)` |
|
| 39972. |
When temperature of a body made of brass increases its moment of inertia. |
|
Answer» increases |
|
| 39973. |
Two wires of same material have masses in the ratio 3:4 the ratio of their extensions under the same load if their lengths are in the ratio 9:10 is |
|
Answer» `5:3` |
|
| 39974. |
The earth revolves round the sun in an elliptical orbit, its speed is |
|
Answer» GOING on DECREASING continuousely |
|
| 39975. |
State newton three laws and discus theirsignificance |
|
Answer» SOLUTION :Newton'sfirstlaw : Everyobjectcontinuestobe in the stateof restor ofuniformmotionunless thereis externalforceactingon it. Newton'ssecondlaw : Theforceactingon an object is equalto the rateof changeof itsmomenturm. `vec(F ) = (vec(dp))/(DT )` newton'sthirdlaw :Newton'sthirdlasw statesthat foreveryactionthere isan equaland oppositereaction . Singificanceof Newton'slaswof MOTION (i)Newton'slawsare vectorlaws: (ii)Theaccelerationexperieneed bythe bodyat time1 dependson theforcewhichactson thebodyat thatinstant oftime (III)In generalthe DIRECTION of a forcemaybe differentfrom the directionof motion (iv) The totalforce `(vec(F )_(max))` is equivalentto thevectorialsum of the individualforces . (v)If noforceactson thebodythenNewton'ssecond`m(d vec(v ))/(dt )=0` (vi) Newton'ssecondlaw has causeadn effectrelation.Force is tehcauseadn accelerationis teheffect. |
|
| 39976. |
If f=x^(2) then what will be relative error in measurement of f ? |
| Answer» SOLUTION :`(DELTAF)/(f)=2(DELTAX)/(X)` | |
| 39977. |
A ball of mass 0.4 kg moving with a uniform speed of 2ms^(-1) strikes normally a wall and rebounds. Assuming the collision to be elastic and the time of contact of the ball with the wall as 0.4s, find the force exerted on the ball. |
|
Answer» 1N |
|
| 39978. |
A body on the surface of the earth at latitude 60° will be under weight lessness state if the angular velocity of rotation of the earth becomes |
|
Answer» `sqrt((GM)/R^(3))` |
|
| 39979. |
What change of surface energy will be noticed when a drop of radius R splits into 1000 droplets radius r, surface tension being T |
|
Answer» `4piR^2T` |
|
| 39980. |
Find the time at which ratio of kinetic energy to potential energy of a particle is 3. The particle starts SHM from its equilibrium position. T is the time period of oscillation: |
|
Answer» `T//6` |
|
| 39981. |
A uniform cube is subjected to volume compression. If each side is decreased by 1%, then bulk strain is: |
|
Answer» 0.01 `V=L^(3)` `RARR""(dV)/(V)=3(dL)/(L)` `THEREFORE%"CHANGE in volume"=3xx(%"change in length")` `=3xx1%=3%` `therefore"Bulk strain"(DeltaV)/(V)=0.03` |
|
| 39982. |
Vectors which act along the same line whose angle between them can be 0^(@) or 180^(@) are called ___________. |
|
Answer» ORTHOGONAL vectors |
|
| 39983. |
Galileo , inhis book Two new sciences , stated that for elevations which exceed or fall short of 45^(@) by equal amounts , the ranges are equal . Prove this statement. |
|
Answer» Solution :For a projectilelaunched with velocity `v_(@)` at an angle `theta_(@)` , therange is given by `R=(v_(0)^(2)sin2theta_(0))/(g)` Now , for angles , `(45^(@)+alpha)and(45^(@)-alpha),2theta_(0)` is`(90^(@)+2alpha)and(90^(@)-2alpha)`, respectively . The valuesof `sin(90^(@)+2alpha)andsin(90^(@)-2alpha)` are the same , EQUAL to that of `cos2alpha`. THEREFORE , RANGES are equal for elevations which exceed of fall short of `45^(@)` by equal amounts `alpha`. |
|
| 39984. |
A ball is projected from a cliff of height h = 19.2 m at an angle a to the horizontal. It hits an incline passing through the foot of the cliff, inclined at an angle theta to the horizontal. Time of flight of the ball is T = 2.4 s. Foot of the cliff is the origin of the co-ordinate system, horizontal is x direction and vertical is y direction (see figure). Plot of y co-ordinate vs time and y component of velocity of the ball (v_(y)) vs its x co-ordinate (x) is as shown.x and y are in m and time is in s in the graph. [g = 10 m//s^(2)] (a) Find the angle of projection a (b) Find the inclination (theta) of the incline. (c) If the ball is projected with same speed but at an angle theta (= inclination of incline) to the horizontal, will it hit the incline above or below the point where it struck the incline earlier? |
|
Answer» |
|
| 39985. |
A uniform cylinder of diameter 8 cm is kept on a rough inclined plane whose angle of inclination with the ground is 30^(@). What should be the maximum height of the cylinder so that it does not topple? |
|
Answer» Solution :The CYLINDER ABCD is kept on the plane of inclination `30^(@)`. Suppose the cylinder is on the verge of toppling over when its HEIGHT is 2 h. At this stage the line of action of gravity MUST pass vertically through the end point A at the base of the cylinder.From Fig tan`30^(@) =(AE)/(EG)` or, `(1)/(sqrt(3))=(r)/(h) " ""or", h = sqrt(3)r = 4sqrt(3) `cm Hence, maximum permissible height = 2h = `2xx4 sqrt(3)` =`8sqrt(3)` cm.
|
|
| 39986. |
A cylinder containe 20 kg of N_(2) gas (M= 28 kg K^(-1) mol^(-1)) at a [ressire pf 5 atm. The mass of hydrogen (M = 28 kg K^(-1) mol^(-1)) at a pressure of 3 atm contained in the same cylinder at same temperature is |
|
Answer» <P>1.08 kg V and T for both CASES in same. Hence `(n_(1))/(p_(1))=(n_(2))/(p_(2))or (m_(1))/(p_(1)M_(1))=(m_(2))/(p_(2)M_(2))` or `""m_(2)(p_(2)M_(2))/(p_(1)M_(1)).m_(1)=((3)(2))/((5)(28)).20=0.86kg` |
|
| 39987. |
A mass of 2 kg oscillates on a spring with constant 50 N/m. By what factor does the frequency of oscillation decrease when a damping force with cosntant b=12 is introduced? |
|
Answer» SOLUTION :`W=SQRT(50/2)=5r//s` `w=sqrt(w^(2)-(b/(2m))^(2))=sqrt(5^(2)-3^(2))=4` fr REDUCES by 1r/s or 20% |
|
| 39988. |
When a satellite is orbitting round a planet in circular orbit, workdone by the gravitational force acting on the satellite is |
|
Answer» ZERO on COMPLETING ONE REVOLUTION only |
|
| 39989. |
Work done to increase the temperature of one mole of an odeal gas by 30^@C, if it is expanding under the condition V prop T^(2//3) is , ( R = 8.314 J// mol//K) |
|
Answer» 16.62 J |
|
| 39990. |
A man is sitting in a boat, which is floating in a pond. If the man drink some water from the pond, will the level of water in the pond decrease ? |
| Answer» SOLUTION :No. When he drinks water, say w kg, he displaces w kg of water and hence the LEVEL TENDS to increase But ALREADY the w kg has gone inside his STOMACH. So the level remains constant | |
| 39991. |
When a constant torque is applied on a rigid body, then |
|
Answer» The body moves with LINCAR acceleration |
|
| 39992. |
A particle undergoes uniform circular motion. The angular momentum of the particle remain conserved about: |
|
Answer» the centre point of the circle `vecL=vecrxxmvecv` `vecL=` Constant when `VECR` and `VECV=` Constant |
|
| 39993. |
Can we predict the direction of motion of a body from the direction of force on it |
| Answer» Solution :Yes. The direction of motion is always opposite to the FORCE of KINETIC friction. By USING the principle of equilibrium, the direction of force of static friction can be DETERMINED. When the object is in equilibrium, the frictional force must point in the direction which RESULTS as a net force is zero. | |
| 39994. |
A particle has an initial velocity 3 hati+4hatj and an acceleration of 0.4 hati + 0.3 hatj. Its speed after 10s is |
|
Answer» 10 UNITS |
|
| 39995. |
why are strings of different thicknesses and materials used in a sitar or a violin ? |
| Answer» SOLUTION :Fundamental frequency of streched string `v= 1/2L sqrt T/m`. when we use string of DEFFERENT thiknesses and materials, they have different values of MASS as PER unit length (m). So the strings will PRODUCED notes of different frequencies. | |
| 39996. |
A particle is projected making an angle of 45^@ with the horizontal having kinetic energy K.The kinetic energy at the highest point will be |
|
Answer» `K/sqrt2` |
|
| 39997. |
The accuracy n. measurement depends on the limit or the resolution of the measuring instrument. State whether the above statement is TRUE or FALSE. |
|
Answer» |
|
| 39998. |
Why are stationary waves called so ? |
| Answer» Solution :In a STATIONARY wave, the PARTICLES of the medium vibrate about their MEAN POSITIONS, but distrbances do MOT treval in any direction. | |
| 39999. |
If S_(P) and S_(V) denote the specific heats of nitrogen gas per unit mass at constant pressure and constant volume respectively, then |
|
Answer» `s_p-s_V=28R` For diatomic gas `N_2` no. of DEGREES of FREEDOM = 5 `S_P - S_V=R/28` |
|
| 40000. |
A body weighs 64 N on the surface of the Earth. What is the gravitational force on it (in N) due to the Earth at a height equal to one-third of the radius of the Earth? |
|
Answer» Solution :LINEAR velocity v=72 km/h = `(72 xx 1000)/(60 xx 60) = 20` m/s Initial angular velocity of the wheel `=omega_(0) = v/r = 20/(0.25) = 80` rad/s Final angular velocity =`omega =0` Time =t= 16S Angular ACCELERATION = `alpha` = ? `alpha = (omega-omega_(0))/t = (0-80)/16 = -5 "rad"//s^(2)` Average torque `=tau = lalpha = 5 xx 5 = 25` Nm |
|