Saved Bookmarks
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.
| 13101. |
A pendulum is hung from the rod of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m"/"s^(2) at a distance of 5 m from the mean position. The period of oscillation is........... |
|
Answer» `1s` `20= W^(2) XX 5` `therefore W^(2) =(20)/(5)= 4` `therefore W= 2` `therefore (2pi)/(T)= 2` `therefore T= (2pi)/(2)` `therefore T= pi s`. |
|
| 13102. |
Solve with due regard to significant figures (i) 46.7-10.4 = (ii) (3.0 xx 10^(-8)) + (4.5 xx 10^(-6))= |
|
Answer» SOLUTION :(i) 46.7- 10.04 Here 46.7 has one decimal PLACE, and 10.04 has TWO decimal places. `therefore 46.7-10.04= 36.66` The result should have only one decimal place. The result is 36.7 (II) `3.0 XX 10^(-8) + 4.5 xx 10^(-6)` `=0.03 xx 10^(-6) + 4.5 xx 10^(-6)` `=4.53 xx 10^(-6)` Here `4.5 xx 10^(-6)` has only one decimal place and `0.03 xx 10^(-6)` can have two decimal places. This result should be rounded off to one decimal place. `therefore (3.0 xx 10^(-8)) + (4.5 xx 10^(-6))= 4.5 xx 10^(-6)` |
|
| 13103. |
What is meant by elastic fatigue? |
| Answer» SOLUTION :ELASTIC fatigue is defined as the loss in the strength of a MATERIAL caused DUE to repeated alternating strains to which the material SUBJECTED. | |
| 13104. |
Two bodies are projected at angles 30^(@) and 60^(@) to the horizontal from the ground uch that the maximum heights reached by them are equal. Then (a) Their times of flight are equal (b) Their horizontal ranges are equal (c) The ratio of their initial speeds of projection is sqrt(3) :1 (d) Both take same time to reach the maximum height. Mark the answer is |
|
Answer» If a, B, C and d are correct |
|
| 13105. |
Match the following : {:("List - 1","List - 2"),("a) Surface Tension","e) Gas constant"),("b) Specific heat","f) Areal velocity"),("c) Latent heat","g) Spring constant"),("d) Kinematic viscosity","h) Gravitational potential"):} |
|
Answer» a-e b-g c-h d-f |
|
| 13106. |
Discuss the importance of the streamlined shape of a fish. |
| Answer» Solution :A fish experiences viscous DRAG force while moving through water. The boy of a fish is tapered at the head and at the TAIL, and is compressed at the sides. This is a type of streamlined shape. For this reason, a fish experiences less viscous drag force while swimming through water and the water flowing by the fish follows streamlines. So they are able to control their DIRECTION of motion very EASILY. For the same reason, the shape of aeroplanes, fast moving trains or racing CARS, are made streamlined. | |
| 13107. |
One mole of a certain ideal gas is given 500 cal heat. As a result the temperature rises by 72 K at constant pressure. Find the work performed by the gas, the change of internal energy and the value of gamma |
| Answer» SOLUTION :`5.98 XX 10^2 J,1502J ,1.4` | |
| 13108. |
A body of mass 5 kg starts from the origin with an initial velocity of vecU=(30i+40j)m//s.A constant force of F=(-hati-5hatj)Nact on the body . Find the time in which the y-component of the velocity become zero . |
|
Answer» SOLUTION :`vecu=30i+40j`….(1) `u=u_(X)hati+u_(y)hatj`….(2) `F=-hati-5hatj `….(3) `F=F_(x)hati+F_(y)hatj`….(4) comparing (1) and(2),(3)and (4) we have `u_(y)=40 m//s, F_(y)=-5N` `F_(y)=ma_(y),``a_(y)=-1m//s^(2)` `v_(y)=u_(y)+a_(y)xxt,0=40-1xxt,t=40sec` |
|
| 13109. |
A mild steel wire of length1.0 m andcross-sectional area 0.50xx10^(-2) cm^(-2) is stretched,well within its elastic limit, horizontally between two pillars.A mass of 100 g is suspendedfrom the mid-point of the wire. Calculate the depression at the mid-point. |
|
Answer» Solution :`x=l[(Mg)/(YA)]^(1//3)` Here, `m 2l=1m, l=0.5m`, `A=0.50xx10^(-6) m^(2) , , M=0.1 kg`, `Y=2xx10^(11) N//m^(2) and g=10m//s^(2)` `therefore` DEPRESSION `x=1.074xx10^(-2)m` |
|
| 13110. |
Disc is in pure rolling as shown below, the v_(1)/v_(2) is |
|
Answer» `1:3` |
|
| 13111. |
Is it possible to realize whether a vessel kept under the tap is about to fill with water ? |
|
Answer» Solution :The frequency of the note produced by an AIR column is inversely proportional to its LENGTH. Asthe level of water is the vessel rises, the length of the air column above it decreases. It PRODUCES sound of decreasing frequeney, i.e., the sound becames shorter. From the shrillness of sound. it is possible to realize whether the vessel is FILLED which water. `v_("min") = 11.71 ms^(-1)` |
|
| 13112. |
The power output of a carnot engine is 100 kW. The engine operates between two reservoirs at 30^(@)C and 300^(@)C (i) How much heat is absorbed per hour ? (ii) How much heat is Tost per hour ? |
|
Answer» Solution :`T_(1) = 30^(@)C + 273 = 303 K, T_(2) = 300 + 273 = 573 K` Efficiency `= ETA = ( T_(1) - T_(2))/( T_(1)) = ( 573 - 303)/( 573) = 0. 476` Power output `= W =Q_(1) - Q_(2) = 100 xx 10^(3) (J)/(S)` `eta = (W)/(Q_(1))` `Q_(1) = (w)/(eta) = (10 ^(5))/( 0. 476) ` `= 2.1 xx 10- ^(5) (J)/(s)` Heta absorbed PER hour `= 2.1 xx 10 ^(5) xx 3600` `= 7. 56 xx 10 ^(8) J` Heat lost per second `= Q_(2) = Q_(1) - W` ` = 2.1 xx 10 ^(5) - 10 ^(5) = 1.1 xx 10 ^(5)` Heat lost per hour `= 1.1 xx 10 ^(5) xx 3600 =3.96 xx 10 ^(8) J` |
|
| 13113. |
(A) : When two boats sail parallel in the same direction and close to each other, they are pulled towards each other. (R ) : The viscous drag on a spherical body moving with speed upsilon is proportional to upsilon. |
|
Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
|
| 13114. |
If the pressure and the volume of certain quantity of ideal gas are halved, then its temperature. |
|
Answer» is doubled |
|
| 13115. |
Simplify the following : (i) (0.0036)^(1//5) (ii) (0.056)^(2//3) (iii) 10^(-1//5) |
|
Answer» `log x =-(1)/(5) log 10 = - (1)/(5)xx1 = - 0.2 = bar1 + 1- 0.2 = bar1.8` Taking antilog on both the SIDES, we GETX = 0.6310 |
|
| 13116. |
A conducting sphere of radius b has a spherical cavity with its centre displaced by 'a' from centre of sphere O_(1). A point charged q is placed at the centre of cavity O_(2). Q charge is given to conducting sphere and charge q_(0) is placed at P, a distance c from centre O_(1). Further O_(1), O_(2) and P are collinear |
|
Answer» CHARGE dispribution on inner surface of cavity is unifrom |
|
| 13118. |
Two projectile A and B located at (0,0) and (4.4) respectively start moving simu ltaneou lsy with velocities V_(A)=-4hati and V_(B)=4hatj: |
|
Answer» the shortest DISTANCE between them is `4sqrt(2)m` Since the 2nd PARTICLE is projected after 2s. Hence the particles will reach the ground in an INTERVAL of 2s. |
|
| 13119. |
A solid whose volume does not change with temperature floats in a liquid. For two different temperatures t_1 and t_2 of the liquid, fractions f_1 and f_2 of the volume of the solid remain submerged in the liquid. The coefficient of volume expansion of the liquid is equal to |
|
Answer» `(f_1 - f_2)/( f_(2) t_(1) - f_(1) t_(2) ) ` |
|
| 13120. |
Mention scalar and vector from g and G. |
| Answer» SOLUTION :`IMPLIES` G is VECTOR while G is SCALAR. | |
| 13121. |
Two identical gas samples are initially at same temperature and pressure. They are separatly taken into two cylinders of volume V (each) and are compressed to volume V/2 One isothermally and other adiabatically.If final pressures in the process are p_1 and p_2 , then the ratio of p_1//p_2is |
|
Answer» `2^gamma` |
|
| 13122. |
Two identical gas samples are initially at same temperature and pressure. They are separatly taken into two cylinders of volume V (each) and are compressed to volume V/2 One isothermally and other adiabatically.Change in entropy of gas |
|
Answer» is more for adiabatic PROCESS |
|
| 13123. |
Two identical gas samples are initially at same temperature and pressure. They are separatly taken into two cylinders of volume V (each) and are compressed to volume V/2 One isothermally and other adiabatically.Entropy change for surroundings is |
|
Answer» POSITIVE for ADIABATIC PROCESS |
|
| 13124. |
Zeroth law of thermodynamics is related to |
|
Answer» WORK |
|
| 13125. |
A large cylindrical tank has a hole of area A at its bottom. Water is poured in the tank by a tube of equal cross-sectional area A ejecting water at the speed upsilon. |
|
Answer» The water level in the tank will KEEP on rising |
|
| 13126. |
A car moving in a straight line at a speed of 120 km covers in 2 hours. There after from that left side it covers 50 km in 1 hours. At that moment its average speed and aveerage velocity will ……. And …… |
|
Answer» Solution :AVERAGE SPEED `= 56.57 KM h^(-1) and ` Average VELOCITY `=43.33 km h^(-1))` |
|
| 13128. |
If 16 gm of oxygen and x gm of hydrogen have the same volume at the same temperature and pressure, then find the value of x. |
|
Answer» Solution :Mass of the oxygen gas `(m_(1))=16` GM ,Mass of th hydrogen gas `(m_(2))=x` From the PERFECT gas equation, PV = nRT `RARR PV=m/M""RT` Since from the given data, pressure volume and temperature are constant, `m/M` constant `therefore m_(1)/M_(1)=m_(2)/M_(2)" where "M_(1)" and "M_(2)` are the molecular weighs of oxygen and hydrogen `16/32=x/2, x=1gm` |
|
| 13129. |
When a bady undergoes an oblique projection, maximum rangeof it is : |
|
Answer» `(U^(2))/(g)` Here `theta = 45^(@)` `R = (u^(2))/(g) sin 90^(@) = (u^(2))/(g)` |
|
| 13130. |
A cyclic process 1rarr2rarr3rarr1 shown in P-T diagram is performed with a constant mass of an ideal gas. Show it on a (a) P-V diagram (b) V-T |
|
Answer» <P> Solution :  (a) `1rarr2` is an inclined line passing through origin i.e., it represents isochoric process (V: constant) `1rarr2` P is increasing `2rarr3`: Pressure is constant `PV=nRTimpliesVpropT` As TEMPERATURE is decreasing, HENCE, volume is decreasing. `3rarr1` Temperature is constant `P-V` diagram will be rectangular hyperbola. In `3rarr1`, pressure is decreasing The `P-V` diagram will be (B) `1rarr2`: Volume is constant, temperature is increasing `2rarr3`: P is constant , on `V-T` diagram `2rarr3` will straight line through origin. T is decreasing. `3rarr1`: T is constant, P is decreasint i.e., volume is increasig (since `PV=nRT`) |
|
| 13131. |
What happens to the value of the force constant of a spring, if the mass attached to it is doubled ? |
| Answer» SOLUTION :The FORCE CONSTANT of the spring is independent of the mass ATTACHED to it. Therefore it does not change with the mass attached. | |
| 13132. |
A satellite is moving around the Earth with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1% its speed will ........ |
|
Answer» INCREASE by 1% |
|
| 13133. |
A solid spherical ball having density d and volume v floats on the interface of two immiscible liquids. The density of the liquid in the upper portion is d_(1) and that of the liquid in the lower portion is d_(2). What parts of the ball will remain in the liquids in the upper and lower portions respectively, if d_(1)ltdltd_(2)? |
|
Answer» Solution :SUPPOSE a volume x of the ball remains in the liquid in the UPPER PORTION. So, a volume (V -x) remains in the liquid in the lower portion. According to the principle of floatation, weight of the body = weight of displaced liquid or, `vd=xd_(1)+(v-x)d_(2)or,x(d_(1)-d_(2))=v(d-d_(2))` or, `x/v=(d-d_(2))/(d_(1)-d_(2))=(d_(2)-d)/(d_(2)-d_(1))` `therefore` The PART of the ball that remains in the liquid in the upper portion = `(d_(2)-d)/(d_(2)-d_(1))` , and the part of the ball remains in the liquid in the lower portion = `(1-(d_(2)-d)/(d_(2)-d_(1)))=(d-d_(1))/(d_(2)-d_(1))`. |
|
| 13134. |
A stone is projected vertically up to reach a maximum height h. The ratio of its kinetic to potential energies at a height (4h)/5 will be |
|
Answer» `5:4` |
|
| 13135. |
A force of 1N acts on a 1kg mass at rest for 1s. In another case 1N force acts on 1kg mass at rest and move it through 1m. The ratio of kinetic energies in the two cases is |
|
Answer» `1 : 1` |
|
| 13136. |
Fill in the blanks using the word from the list appended with each statement. The working of a hydraulic lift is based on cdots (Pascal's law /principle of Conservation of Energy) |
|
Answer» |
|
| 13137. |
The CM of a uniform card board cut in .T. shape as shown in figure is : |
|
Answer» 4 CM from A TOWARDS B |
|
| 13138. |
Two masses 10gm and 40gm are moving with kinetic energies in the ratio 9: 25. The ratio of their linear momenta is |
| Answer» Answer :B | |
| 13139. |
The velocity - time graph of an object is given below: The area under this graph gives |
| Answer» SOLUTION :DISPLACEMENT | |
| 13140. |
In the diagram shown in figure. Match the following columns. {:("Colmun I","Column II"),("(A) Absolute acceleration of 1 kg block",(p) 11 ms^(-2)),("(B) Absolute acceleration of 2 kg block",(q)6 ms^(-2)),("(C) Relative acceleration between the two",(r) 17 ms^(-2)),(,(s) "None"):} |
Answer» 
|
|
| 13141. |
When the pressure acting on an oil of volume 0.5 lit is increased by 15 xx 10^(5)N m^(2) the change in volume is 0.3 cm^(3). Find the bulk modulus of oil. |
|
Answer» Solution :Volume strain `=("change of volume")/("original volume")=(0.3)/(500)=3//5000` STRESS = Pressure change `=15xx10^(5)Nm^(2)` `:.` BULK MODULUS `=("stress")/("strain")=(15xx10^(5))/(3//5000)=25xx10^(8)N//m^(2)` So bulk of the oil `=0.25xx10^(10)N//m^(2)` |
|
| 13142. |
Why calorimeters are made of metal (mostly copper) instead of glass? |
|
Answer» Solution :Due to higher conductivity of metals, in a metallic calorimeter thermal equilibrium is achieved very fast. Also specific heat of metals is lower than that of glass. That is why the water equivalent of a metallic calorimeter is lower than that of a glass calorimeter of the same mass. Therefore, for the same rise in temperature heat gained by metallic calorimeter is LESS than that gained by glass calorimeter. As a result the final temperture of the mixture will be higher and will result less error in measurement. COPPER has the highest conductivity (other than silver) AMONG all the metals, and copper is much cheaper than silver, that is why copper is mostly used as the material of a calorimeter. VALUE of t can be calculated from this equation if values of all the other quantities are known. Thus, MEASURING temperatures `t_(1) and t_(2)` with the help of a thermometer graduated up to `100^(@)C`, a much greater temperature t can be determined. |
|
| 13143. |
There is a ring or mass m and radius R is pivoted at a point O on its periphery. It is free to rotate about an axis perpendicular to its plane. What is the period of ring? |
|
Answer» `T=2pi(((R)/(g)))` `because""T=2pisqrt([(l)/(MGD)])T=2pisqrt([(2mR^(2))/(MGR)])` `rArr""T=2pisqrt([(2R)/(g)])` 
|
|
| 13144. |
The mean radius of earth is R, its angular speed on its own axis is omegaand the acceleration due to gravity at earth.s surface is g. What will be the radius of the orbit of a geostationary satellite ? |
|
Answer» `(R^(2)G//OMEGA^(2))^(1//3)` |
|
| 13147. |
The radius of acceleration due to gravity at a height 1 km above and 1km below the earth surface is [R -radius of the earth ] |
|
Answer» `(R-2)/(R-1)` |
|
| 13148. |
The accuracy in the measurement of speed of light is 3.00xx10^(8)m//s is |
|
Answer» `(1)/(300)%` |
|
| 13149. |
Two simple harmonic motions are represented by the equations y_(1)=0.1sin(100pit+pi/3)andy_(2)=0.1cospit. Find the initial phase difference of the velocity of particle 1, with respect to the velocity of particle 2. |
|
Answer» |
|
| 13150. |
The height at which the acceleration due to gravity becomes g/9(where g = acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is ......... |
|
Answer» `sqrt2R` `g/9 = (GM_e)/((R+H)^2)` `:. (GM_e) /(9R^2) =(GM_e)/((R+h)^2) ""[ :. g = (GM_e)/R^2]` `:. 9R^2 = (R+h)^2` `:. 3R = R + h ` `:. h = 2R` |
|