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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.
| 13401. |
A stationary wheel starts rotating about its own axis at constant angular acceleration. If the wheel completes 50 rotations in first 2 seconds, then the number of rotations made by itin next two seconds is |
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Answer» 75 |
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| 13402. |
Light year is the unit of |
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Answer» velocity |
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| 13403. |
Find the minimum velocity for which the horizontal range of a projectile is 39.2 m. |
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Answer» |
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| 13404. |
What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity ? |
| Answer» Solution :FREQUENCY `v =1/(2pi)sqrt(G/l)` .The value of g inside the freely falling cabin is ZERO. So frequency is zero. | |
| 13405. |
The velocity with which a gun suddenly moves backward after firing is |
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Answer» LINEAR VELOCITY |
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| 13406. |
The planet mercury is revolving in an Elliptical orbit around the sun as shown in the figure. The kinetic energy of mercury will be greatest at |
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Answer» A |
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| 13407. |
Give some uses of geostationary satellites. |
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Answer» Solution :In communicating radio, T.V and telephone signals across the world. In studying upper regions of the ATMOSPHERE. In FORECASTING weather. In deter ming the exact shape and dimensions of the EARTH. In studying solar RADIATIONS and cosmic rays. |
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| 13408. |
The volume of a bulb of mercury thermometer at 0^(@) C is V_0 and cross sectional area is A_0 and alpha_(g) is the coefficient of linear expansion of the glass and gamma_(m) is the coefficient of cubical expansion of mercury. If the mercury fills the bulb at 0^(@) C what is the length of mercury in capillary at t^(@) C. |
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Answer» `(V_(0) t ( gamma_(m) - 3 alpha_(g)) )/( A_(0) [ 1+ 2 alpha_(g) t ] )` |
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| 13409. |
A cube with a mass m = 20g wettable by water floats on the surface of water. Each face of the cube is a = 3cm long. Surface tension of water is 70 dyne/cm. The distance of the lower face of the cube from the surface of water is (x+0.3)cm. Find x. (g = 980cm/s^2) |
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Answer» |
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| 13410. |
An accurate Celsius thermometer and Fahrenheit thermometer register 60° and 141 ° respectively when placed in the same constant temperature enclosure. What is the error in the Fahrenheit thermometer? |
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Answer» `1^(@) F` |
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| 13411. |
Where will a person hear maximum sound, at node or anti-node? |
| Answer» SOLUTION :Perception of SOUND is DUE to pressure variations that is MAXIMUM at NODES. | |
| 13412. |
Find the value of k in equation in day and kilometer. k = 10^(-13)s^(2)m^(-3)The distance of artificial satellite from earth is 14R. Where Re = radius of earth. Find its time period. |
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Answer» |
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| 13413. |
We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron (beta_("vbrass")=6xx10^(-5)//K" and "beta_("viron")=3.55xx10^(-5)//K) to create a volume of 100 cc. How do you think you can achieve this. |
Answer» Solution :Let, `V_(io)` and `V_(bo)` are volume of iron and brass at `0^(@)C` respectively.  `V_(i),V_(b)` are volume of iron and brass respectively of `DeltaT^(@)C`. `gamma_(i)` and `gamma_(b)` are coefficient of volume EXPANSION of them respectively. According to question, `V_(io)-V_(bo)=V_(i)-V_(b)=100" cc"` . . .(1) `:.V_(i)=V_(io)(1+gamma_(i)DeltaT)` and `V_(b)=V_(bo)(1+gamma_(b)DeltaT)` `:.` From EQUATION (1), `V_(i)-V_(b)=V_(io)(1+gamma_(i)DeltaT)-V_(IB)(1+gamma_(b)DeltaT)` `V_(i)-V_(b)=V_(io)-V_(ib)+V_(io)gamma_(i)DeltaT-V_(ib)gamma_(b)DeltaT` But `V_(i)-V_(b)=V_(io)-V_(ib)` `:.V_(io)gamma_(i)DeltaT=V_(ib)gamma_(b)DeltaT` `:.(V_(io))/(V_(bo))=(gamma_(b))/(gamma_(i))` for midpoint `=(6xx10^(-5))/(3.55xx10^(-5))=(6)/(3.55)` `:.V_(io)=(6)/(3.55)V_(bo)`. . . (2) `:.` From equation `V_(io)-V_(ib)=100`, `(6)/(3.55)V_(bo)-V_(bo)=100` `:.6V_(bo)-3.55V_(bo)=355` `:.2.45V_(bo)=355` `:.V_(io)=144.89" cc"` `:.V_(io)~~144.9" cc"` Now, again `V_(io)-V_(bo)=100` `:.144.89-V_(bo)=100` `:.144.89-100=V_(bo)` `:.44.89=V_(b)` `:.V_(bo)~~44.9" cc"` |
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| 13414. |
If radius of earth is R, then the height h at which the value of 'g' becomes one fourth is ................ |
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Answer» Solution :`implies g/4 = (GM_e)/((R_e+h)^2)` `:. 1/4 ((GM_e)/R_e^2)=(GM_e)/((R_e+h)^2)` `:. (R_e +h)^(2)= 4R_e^2` `:. R_e+ h = 2R_e` ` :. R_e = h` TAKING `R_e` = RADIUS of earth =R h = R |
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| 13415. |
If the linear ensity of a rod of length L varies as lambda = A+Bx, find the position of its centre of mass. |
Answer» Solution :Let the x-axis be along the length of the ROD and ORIGIN at ONE of its ends. As rod is along x-axis, for all points on it y and Z co - ordinates are zero. Centre of mass will be on the rod. Now CONSIDER an element of rod of length dx at a distance x from the origin, then `dm = lambda dx = (A+Bx)dx` `X_(CM)=(int_(0)^(L) xdm)/(int_(0)^(L)dm)=(int_(0)^(L)x(A+Bx)dx)/(int_(0)^(L)(A+Bx)dx)` or `X_(CM)=((AL^(2))/(2)+(BL^(3))/(3))/(AL+(BL^(2))/(2))=(3Al+2BL^(2))/(3(2A+BL))=(L(3A+2BL))/(3(2A+BL))` |
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| 13416. |
In a circus show, a motorcycle rider inside a ' deathwell' can revolve on the erect wall without falling down. What is the reason behind it? |
| Answer» Solution :While revolving on the erect wall inside the well, the rider derives the necessary centripetal force for revolving in a circular PATH from the horizontal component of normal force. Due to the presence of this normal force, an upward FRICTIONAL force is GENERATED which balances the combined weight of the motor-cycle and the cyclist. For this REASON, the rider does not FALL down. | |
| 13417. |
Which of the following sets of forces acting simultaneously on a particle keep it in equilibrium? (in newton) |
| Answer» Answer :C | |
| 13418. |
A piece of metal weighs 46g times g in air. It weighs 30g times g in a liquid of specific gravity 1.24 at 27^(@)C. At 42^(@)C, when the specific gravity of the liquid is 1.20, the weight of the piece immersed in it is 30.5g times g.Find the coefficient of linear expansion (alpha) of the metal. |
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Answer» SOLUTION :Mass of the displaced liquid at `27^(@)C` `""=46-30=16g` Volume of the displaced liquid at `27^(@)C, V_(27)=16/1.24 cm^(3)` Similarly volume of the displaced liquid at `42^(@)C`, `""V_(42)=(46-30.5)/1.20=15.5/1.20cm^(3)` So the volume of the PIECE at `27^(@)C " and " 42^(@)C` are `16/1.24cm^(3) " and " 15.5/1.20cm^(3)` respectively. Now, `V_(42)=V_(27){1+gamma(42-47)}` [where `gamma`= COEFFICIENT of volume expansion of the metal] or, `""15.5/1.20=16/1.24{1+gamma times 15} " or, " 1+15gamma=(15.5 times 1.24)/(1.20 times 16)` or, `"" 15gamma=1.001-1` or, `"" gamma=0.001/15=3alpha` `therefore` Coefficient of linear expansion of the metal, `"" alpha=gamma/3=0.001/45=2.2 times 10^(-5@)C^(-1).` |
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| 13419. |
A force represented as shown in figure acts on a body of mass 16 kg, the body starts from rest. |
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Answer» the change in MOMENTUM ofduring the first 5 second of its motion is 250 N - S |
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| 13420. |
Compared to a burn due to water at 100^@C , a burn due to steam at 100^@Cis |
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Answer» More DANGEROUS |
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| 13421. |
A gun fires 8 bullets per second into a target X If the mass of each bullet is 3 g and its speed 600ms^(-1). Then, calculate the power delivered by the bullets |
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Answer» Solution :`P=8xx("Kinetic ENERGY of each bullets PER SECOND")` `=8xx1/2xx(3xx10^(-3))xx(600)^(2)=4320W , P=4320kW` 
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| 13422. |
Answerthe following question based on the P - T phase diagram of CO_2 as given in Q. 16 (a) CO_2 at 1 atm. Pressure and temperature-60^@C is compressed isothermally. Does it go through a liquid phase ? (b) What happens when CO_2 at 4 atm. pressure is cooled from room temperature at constatn pressure ? (c ) Describe qualitatively the changes in a given mass of solid CO^@ at 10 atm. pressure and temperature-65^@C as it is heated up to room temperature at constant pressure. (d) CO_2 is heated to a temperature 70^@C and compressed isothermally. What change in its properties do you expect to observe ? |
| Answer» Solution :(a) SINCE the temperature ` -60^@V` lies to the left of `56.6^@C` on the curve i.e. lies in the region of vapour and solid phase, so carbon dioxide will condens directily into solid without becoming liquid. (b) Since the pressure 4 ATM. Is less then 5.11 atm. the carbon dioxide will condense directly into solid without becoming liquid. (c ) When a solid `CO_2`at 10 atm. pressure and `-65^@C` temperature is heated it is first converted into a liquid. A further increases in temperature brings it into the vapour phase. At = P = 10 atm, if a horzontal line is drawn parallel to the T -axis , then the points of intersection of this line with the fusion and vaporization curve will give the fusion and BOILINGPOINTS of `CO_2` at 10 atm. (d) Since `70^@C` is higher then the critical temperature of `CO_2`, so the `CO_2` gas cannot be converted into liquid state on being compressed isothermally at `70^@C` it will remain in the vapour state. However, the gas will depart more and more form its perfect gas behaviour with the increase in pressure. | |
| 13423. |
A lift is going up, the total mass of the lift and the passenger is 1500 kg. The variation in the speed of the lift is as shown in (a) What will be the tension in the rope pulling the lift at, (i) 1s (ii) 6 s (iii) 11 s (b) What is the height to which the lift takes the passenger? (c) What will be the average velocity and acceleration during the course of the entire motion? (g=9.8m//s^(2)) |
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Answer» Solution :(a) As slope of `v//t` graph gives acceleration, so (i) At `t=1s` `a=(3.6-0)/2=1.8m//s^(2)` i.e, lift is acc. Up, so `T=m(g+a)` i.e, `T=1500(9.8+1.8)` or `T=17400 N` (II) At `t=6` s `a=(3.6-3.6)/(10-2)=0m//s^(2)` i.e., lift has no acc,, so `T=mg` i.e, `T=1500xx9.8` or `T=14700N` (iii) At `t=11` s `a=(0-3.6)/(12-10)=-1.8m//s^(2)` i.e, lift is acc. down, so `T=m(g-a)` i.e, `T=1500(9.8-1.8)` or `T=12000N` (b) The displacement of the passenger is equal to the area under `v//t` curve i.e,  `=1/2(2-0)xx3.6+(10-2)xx3.6+1/2(12-10)xx3.6` `=3.6[1+8+1]=36m` `"AVERAGE velocity"=("displacement")/("time TAKEN")` `=36/12=3m//s` `"and Average acceleration"=("CHANGE in velocity")/("time taken")` `=(0-0)/12=0` |
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| 13424. |
A metal rod has a length of 1m at 30°C. 'alpha' of metal is 2.5 xx 10^(-3) //^(@) C. The temperature at which it will be shortened by 1 mm is |
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Answer» `- 30^(@) C` |
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| 13425. |
If the displacement of object moving in straight line proportional to square of time, then what will be constant ? Velocity of acceleration ? |
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Answer» Solution :CONSTANT acceleration, `x prop t ^(2)` `V = (dx)/(dt) IMPLIES v prop t` ` a = (DV)/(dt) implies a =` constant |
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| 13426. |
Particle A starts from rest and moves along a straight line. Acceleration of the particle varies with time as shown in the graph. In 10 s the velocity of the particle becomes 60 m//s and the acceleration drops to zero. Another particle Bstarts from the same location at time t = 1.1 s and has acceleration – time relationship identical to A with a delay of 1.1 s. Find distance between the particles at time t = 15 s. |
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Answer» |
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| 13428. |
A liquid having coefficient of viscosity n flow through a tube of length at the rate of 100 cc per sec. The pressure difference between the entering and leaving end of tube is equivalent to p. A second tube of radius (r)/(2) but of same length is connected in series with first tube and is connected to the same source. Rate of pressure difference across 1 and 2 tube when they are connected in series is |
| Answer» ANSWER :D | |
| 13429. |
The displacement of a body is given byx = 2t^(3) - 6t^(2) + 12t + 6. The acceleration of body is zero at time ti is equal to : |
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Answer» 3s `v = (dx)/(dt) = (d)/(dt) (2t^(3) - 6t^(2) +12t +6)` `= 2 XX 3t^(2) - 6 xx 2t +12` `= 6t^(2) - 12t + 12` `a = (dv)/(dt) = (d)/(dt) (6t^(2) - 12t+12)` `= 6 xx 2t - 12` `= 12t - 12` Sincea = 0 12t - 12 = 0 12t = 12 `:. t = (12)/(12) = 1s` t = 1s |
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| 13430. |
Assertion :As the mass of simple pendulum increases, its periodic time increases. Reason : Periodic time of simple pendulum is T= 2pi sqrt((l)/(g)). |
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Answer» Both are ture and the REASON is the CORRECT EXPLANATION of the ASSERTION. |
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| 13431. |
A liquid having coefficient of viscosity n flow through a tube of length at the rate of 100 cc per sec. The pressure difference between the entering and leaving end of tube is equivalent to p. A second tube of radius (r)/(2) but of same length is connected in series with first tube and is connected to the same source. If the two tubes are connected in parallel then the new rate of flow will be |
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Answer» `(100)/(17)` CC / sec |
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| 13432. |
A liquid having coefficient of viscosity n flow through a tube of length at the rate of 100 cc per sec. The pressure difference between the entering and leaving end of tube is equivalent to p. A second tube of radius (r)/(2) but of same length is connected in series with first tube and is connected to the same source. Initial flow is given as 100 cc/sec. then final flow will be |
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Answer» `(100)/(17)` cc / SEC |
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| 13433. |
Due to heating, soft rubber |
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Answer» EXPANDS |
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| 13434. |
Why bottom filled with liquid breakdown when cork is hit ? |
| Answer» SOLUTION :The pressure according to `(F)/(a)` is produced when cork of area .a. HITS .The pressure spreads in whole liquid according to Pascal.s LAW the area of bottom is much greater hence FORCE F=PA exerted which is the CAUSE of bottle breakdown. | |
| 13435. |
A block of mass4 kg is pushed against a rough vertical wall with a force 80 N is applied on the block in a direction parallel to the wall . Will the block move? If yes , what is acceleration of the block ? |
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Answer» Solution :`N_(0) =80 N` `(f_(s))_(max) = mu_(s) N = 0.5 xx 80 = 40 N` `F : resultant force = SQRT((30)^(2) + (40)^(2)) = 50 N` `tan theta = (30)/(40) = (3)/(4) rArr theta = 37^(@)` The BLOCK will move when the resultant external force on it exceeds `(f_(s))_(max)`. When it moves , the direction of the force of friction will be opposite to the block's direction of motion `a = (F - (f_(s))_(max))/(m) = ( 50 - 40)/(4) = 2.5 m//s^(2)`   
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| 13436. |
Infinte numberof conducting rods eachof length Land thermalconductivityK arearranged as shown. The cross rods are of crosssectionA, whilethe remainingones are of cross section 2A each . The twoends P and Q are maintained at temperatureT_(1)and T_(2) (T_(2) gt T_(1)) . Find the amountof heatflowingper second throughthe systemof rods . |
| Answer» Solution :`(2(T_(2) - T_(1))AK)/(sqrt(5)L)` | |
| 13437. |
What are the different types of vectors? |
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Answer» SOLUTION :1.EQUAL vectors: Two vectors `vecAandvecB` are said to be equal when they have equal magnitude and same direction and represent the same physical quantity (a) Collinear vectors: Collinear vectors are those which act along the same line. The ANGLE between them can be `0^(@)` or `180^(@)`. (i) Parallel vectors: If two vectors A and B act in the same direction along the same line or on parallel lines, then the angle between them is `0_(0)`.  (ii) Anti-parallel vectors: Two vectors `vecAandvecB` are said to be anti-parallel when they are in opposite directions along the same line or on parallel lines. Then the angle between them is `180^(@)`.  Unit vector: A vector divided by its magnitude is a unit vector. The unit vector for `vecA` is denoted by `hatA`. t has a magnitude equal to unity or one. Since, `hatA=(vecA)/(A)`, we can write `vecA=AhatA` Thus, we can say that the unit vector SPECIFIES only the direction of the vector quantity  3 Orthogonal unit vectors: Let `hati,hatjandhatk` be three unit vectors which specify the directions along positive x-axis, positive y-axis and positive z-axis respectively. These three unit vectors are directed perpendicular to each other, the angle between any two of them is `90^(@).hatihatjandhatk` are examples of orthogonal vectors. Two vectors which are perpendicular to each other are called orthogonal vectors as shown in the figure. 
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| 13438. |
(A) : The magnitude of a physical quantity does not change when the system of units is changed from S.I system to C.G.S system. (R) : The magnitude of a physical quantity is independent of system of units |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 13439. |
One of the faces of a copper cube of side 7.7 cm is maintained at 100°C and the opposite face at 30°C. If the thermal conductivity of copper is 385 "Wm"^(-1) "K"^(-1). Calculate the rate of heat flow through the cube? |
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Answer» Solution :`X= 7.7 xx 10^(-2) m` `A = x^(2) = 59.92 xx 10^(-4)m^(2)` `K= 385 " Wm"^(-1) "K"^(-1)` `theta_1 - theta_2 = 100-30 = 70^(@) C = 70 "K"` `t= 1 s` Quantity of heat FLOWING is one second `= "KA" ( theta_1 - theta_2) t//x` `=( 385 xx 59.29 xx 10^(-4) xx 70 xx 1)/( 7.7 xx 10^(-2) ) = 2075.15` J/S 
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| 13440. |
Why building get demolished by earthquakes? |
| Answer» SOLUTION :An EARTHQUAKE induces FORCED vibration in the WALLS of buildings. For intense earthquakes, the amplitudes of the forced vibrations become very high, which in manycases cross the elastic limit of the constituent materials of the buildings. So MANY buildings get demolished. | |
| 13441. |
The pilot of an aeroplane flying horizontally at a height of 2000m with a constant speed of 540 kmph wishes to hit a target on the ground . At what distance from the target should release the bomb to hitthe target ? |
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Answer» Solution :Given : Initialof the bomb in the horizontal is the same as that of the aeroplane . Initial velcoity of the bomb in the horizontal direction = 540 KMPH ` = 540 xx (5)/( 18) ms^(-1)` `= 150 ms^(-1)` Initial velocity in the vertical direction (u) = 0 `{:("Vertical"),("distance(s)"):}} = 2000 m ` , TIME of FLIGHT t = ? ` s = ut +(1)/(2) at^(2)` Substituting the known values , ` 2000 = 0 xx t +(1)/(2) xx 9 . 8 xx t^(2)` ` 2000 = 4.9 t^(2)` (or) `t = sqrt((2000)/( 4 . 9)) = 20 . 20 s` `:.{:("Horizontal"),("range"):}} = {{:("Horizontal velocity"),(xx "time of flight"):}` ` = 150 xx 20 . 20 = 3 030 m ` 
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| 13442. |
The triple points of neon and carbon dioxide are 24.57 K and 216.55 respectively. Express these temperatures on the Celsius and Fahrenheit scales. |
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Answer» Solution :(1) TRIPLE point for Neom `T=24.57` K `:.T_(C)=T(K)-273.15` `=24.57-273.15` `=-248.58^(@)C` Now, `T_(F)=9/5T_(C)+32` `=9/5(-248.58)+32` `=-447.44+32` `=-415.44^(@)F` (2) Triple point for `CO_(2),T=216.55K` `T_(C)=T(K)-273.15` `=216.55-273.15` `=-56.6^(@)C` `:.T_(F)=9/5T_(C)+32` `=9/5(-56.6)+32` `=-101.88+32` `=-69.88^(@)F` |
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| 13443. |
Give the characteristics of shm. |
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Answer» Solution :(i) The motion is to and fro and periodic. (ii) The particle is under a restoring force proportional to the displacement from equilibrium position and always directed towards the equilibrium position. (iii) Velocity of the particle when its displacement is x is `V = omegasqrt(a^2-x^2)` . THUS velocity is MAXIMUM at the equilibrium position and zero at the extreme position. (iv) Acceleration of the particle is `a = - omega^2x`. Acceleration is maximum acceleration is maximum at the extreme position and zero at the equilibrium position. (v) At the equilibrium position the K.E. is maximum, P.E. is zero. At the extreme position, the K.E. is zero, P.E. is maximum, but the TOTAL energy is always a constant equal to `1//2 ka^2`where k is the force constant and .a. is the AMPLITUDE. |
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| 13444. |
Kinetic energy of a lighter body is equal to that of a heavier body. Which one ofthem has a greater momentum ? |
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Answer» Solution :Let mass and velocity of the LIGHTER body be m and v,and those of the heavier body be M andV , respectively. GIVEN `1/2 MV^2 =1/2 mv^2 or, (M^2V^2)/(M)=(m^2v^2)/(m)` `or, (m^2v^2)/(M^2V^2) =m/Mor, (mv)/(MV) = sqrt(m/M)` or, `("MOMENTUM of the lighter body")/("momentum of the heavier body") = sqrt(m/M) lt 1` HENCE, momentum of the heavier body is greater. |
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| 13445. |
Straws are used to take soft drinks . Why ? |
| Answer» Solution :Straws are USED because when we suck through the STRAW , the pressure inside the straw becomes less than the atmospheric pressure .Due to the DIFFERENCE in pressure the SOFT drink RISES in the straw and we are able to enjoy it conveniently . | |
| 13446. |
Two paper screens A and B are separated by a distance of 100 m. A bullet pierces A and then B. The hole in B is 10 cm below the hole in A. If the bullet is travelling horizontally at the time of hitting the screen A, calculate the velocity of the bullet when it hits the screen A. Neglect the resistance of paper and air. |
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Answer» Solution :The situation is shown in fig. `d = usqrt((2(h_(1)-h_(2)))/(G))` `100 = u SQRT((2 XX 0.1)/(9.8))` `u = 700 m//s` 
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| 13447. |
Two moles of monoatomic gas is expanded from (P_0, V_0)to (P_0, 2V_0)under isobaric condition. Let Delta Q_1, be the heat given to the gas, DeltaW_1 , the work done by the gas and Delta U_1 , the change in internal energy. Now the monoatomic gas is replaced by a diatomic gas. Other conditions remaining the same. The corresponding values in this case are DeltaQ_2,Delta W_2, Delta U_2respectively, then |
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Answer» `DELTA Q_1 - Delta Q_2 = Delta U_1- Delta U_2` |
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| 13448. |
A linear harmonic oscillator of force constant 2xx10^(6)Nm^(-1) and amplitude 0.01 m has a total mechanical energy of 160 J. Then find maximum and minimum values of P.E and K.E? |
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Answer» Solution :MAXIMUM `K.E= (1)/(2)KA^(2)= (1)/(2)xx2xx10^(6)xx(0.01)^(2)= 100j` Since total energy is 160J, maximum P.E is 160J from this it is UNDERSTOOD that at the mean position potential energy of the simple harmonic oscillator is minimum which need not be ZERO. `PE_("min")= TE-KE_("max")= 160-100= 60J= KE_("min")= 0` |
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| 13449. |
A diatomic ideal gas is heated at constant volume until the pressure is doubled and then heated at constant pressure until volume is doubled. The average molar heat capacity for whole process is |
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Answer» `13/6R` |
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| 13450. |
In a compound microscope the focal length of objective lens is 1.2cm and focal length of eye piece is 3.0cm When object is kept at 1.25cm in front of objective, final image is formed at infinity, Magnifying power of the compound microscope should be |
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Answer» 150 |
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