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InterviewSolution
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.
| 1801. |
An object starting from rest undergoes an acceleration of 8ms ^(-2)for 10 second. Find (a) the speed (b) the distance travelled. |
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Answer» Solution :`u =0, a =8 MS ^(-2), t =10` second `(a) V =u +at` `=0 + 8 ms ^(-2) xx 10s ` `= 80 ms ^(-1) ""..(1)` ` (b) s = ut + (1)/(2) at ^(2)` ` = 0 xx 10 + (1)/(2) xx 8 ms ^(-2) xx (10 s ) ^(2) ` `=0 + 4 xx 100 m = 400 m ""...(2)` The speed of the OBJECT is `80 ms ^(-1),` the distance travelled by the object is 400 m. |
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| 1802. |
An object of mas 12 kg is at a certain height above the ground. If the potential energy of the object is 480 J, find the height at which the object is with respect to the ground. Given g=10ms^(-2). |
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Answer» Solution :Mass of the object `m=12kg`, POTENTIAL energy `E_(p)=480J` `E_(p)=MGH` `therefore 480J=12kgxx10ms^(-2)xxh` `H=(480J)/(120kgms^(-2))=4M` The object is at the height of 4 m. |
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| 1803. |
Is the earth really spherical in shape? |
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| 1804. |
A boy of mass 40 kg climbs a staircase of 30 steps, each of 0.2 m height, in 30 s. Calculate power. (g=9.8ms^(-2)) |
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| 1805. |
Distinguish between speed and velocity. |
Answer» SOLUTION :
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| 1806. |
What do you conclude about the speed of light in diamond if the refractive index of diamond is 2.41? way from the mirror. Does the image become smaller or larger? What do you observe? |
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Answer» SOLUTION :`MU = 2.41 , Ca = 3 xx 10^(8) ms` [ VELOCITY of light ] `mu = ( " Speed of light in AIR")/("Speed of light in diamond" ) = (C_a)/(C_d)` `C_d = (C_a)/( mu ) = (3 xx 10^(8))/(2.41) = 1.245 xx 10^(8)` m/s `therefore` Speed of light decreases when the light ray travels air to diamond. |
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| 1807. |
Define standard unit. |
| Answer» SOLUTION :UNIT is the quantity of a constant magnitude which is used to measure the MAGNITUDES of other QUANTITIES of the same NATURE. | |
| 1808. |
You are provided with a hollow iron ball A of volume 5cm^(3) and mass 12 g and a solid iron ball B of mass 12 g. Both are plced on the surface of water contained in a largetub. A. Find upthrust on each ball. B. Which ball will sink? Give reason for your answer. (Density of iron =8.0gm^(-3)) |
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| 1809. |
A solid of density rho has weight W. Show that its apparent weight will be W[1-(rho_(L)//rho)] when it is completely immersed in a liquid of densilty rho_(L). |
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Answer» Solution :Given WEIGHT SOLID =W `:.` Mass of solid `=W//g` Voluem of solid `=("Mass")/("Density")=(W//g)/(rho)` Volume of liquid displaced = Volume of solid `=(W//g)/(rho)` upthrust on solid = Volume of liquid displaced `xx` density of liquid `X` ACCELERATION due to gravity or UPTHURST `=((W//g)/(rho))xxrho_(L)xxg=W/(rho)xxrho_(L)` `:.` Apparent weight `=` True weight - upthurst `=W-(W/(rho)xxrho_(L))=W(1-(rho_(L))/(rho))` |
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| 1810. |
A body of mass 'm' is kept on the horizontal floor and it is pushed in the horizontal direction with a force of 10N continuously, so that it moves stadily. (a) Draw FBD (a diagram showing all the foreces acting on the body aty a point of time) (b) What is the value of friction ? |
Answer» SOLUTION : Given tht the body is moving steadily, Hence the net force on the body is zero both in horizontal and vertical directions. Forces ACTING on it along horizontal direction are force of friction (f), force of PUSH (F) We know the `f _(N et,X) =0` `F + (-f) =0` `F =f` Hence the vlaue of force of friction is 10N. |
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| 1811. |
How does the water kept in an earthen pot remains cool? |
| Answer» SOLUTION :An earthen pot consists of small PORES from which the water inside the pot constantly seeps out and gets evaporated due to the presence of high temperature AROUND it. The evaporation process REQUIRES heat which is acquired from the surface of the pot, hence MAKING the water and the pot cooler. | |
| 1812. |
A graph of velocity-time (v-t) for an object performing motion is shown, what would you conclude about motion of the object? |
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Answer» It would be performing uniform motion. |
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| 1813. |
If the distance between two bodies that attract each other is doubled, how many times more will be their. mutual force of attraction? |
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| 1814. |
A beaker full of water is suspended from a spring balance. Will the reading of the balance change: If a cork is placed in water ? If a piece of heavy metal is placed in it ? Give the reasone for your answer |
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| 1815. |
The speed of the scooter moving with uniform acceleration of 4 m s^(-2) becomes 20 ms^(-1) at a certain time. What will be the speed of the scooter, when it has covered a distance of 112 m after that time? |
| Answer» SOLUTION :`36 KM H ^(-1)` | |
| 1816. |
A trolley, while going doen an inclined plance, has an acceleration of 2 cm s ^(-2). What will be its velocity 3 s after the start ? |
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Answer» Solution :Here, `U = 0,a =2 CM s ^(-2) ,t =3 s ` `v=u +at` `= 0 + 2 xx 3 = 6 cm s ^(-1)` The velocity of the TROLLEY will be `6 cm s ^(-1).` |
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| 1817. |
A vehicle has a mass of 1500 kg. What must be the force between the vehicle and the road if the vechicle is to be stoped with a negative acceleration of 1.7 ms ^(-2) ? |
| Answer» SOLUTION :2550 N in a direction OPPOSITE to that of the MOTION of the vechcle `(AS _(7))` | |
| 1818. |
For which of the following physical quantities it is necessary to indicate direction along with its magnitude ? |
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Answer» SPEED |
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| 1819. |
When thetip ofthescrew of a screw gauge is incontactwith thestudthe zeroof thecircularscaleis 3 divisionsbelowthe indexline. To holda wire the thimble is given a little over 3rotationsin theanti-clockwisedirection and with wire heldtightly the 32nddivision of the circular scaleis nowin linewith the baseline. Thepitchof the screw is 0.5mm andthe numberif divisions on thecircular scaleare 100. Findthe correct diameter ofthewire. |
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Answer» Solution :(i) `T a .(1)/(sqrt(g))` `(T_(p))/(T_(m)) =sqrt((g_(m))/(g_(p)))` Given , `T_(m)=2 s,g_(m)=6g_(E) " and" g_(p) =2g_(e)` Here thesuffixes pm anderepresent planetmoonand earthrespectively. (ii) 3.45 s |
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| 1820. |
In a Roman steel yard, the distance of the rider from its zero mark is proportional to the |
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Answer» weight of the LOAD. |
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| 1821. |
What are the changes of state in water? Explain. |
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Answer» Solution :Any matter around us can be in the three forms : solid ,liquid and gas called states of matter. Depending upon the temperature, pressure and TRANSFER of heat, matter is converted from one state to another and is known as change of state in matter. There are different such processes in the change of state inmatter. For example: WATER molecules are in liquid state at normal temperature. When water is heated to `100^@C`, it becomes steam or vapour which is a gaseous state of matter. The process by which a liquid is converted to vapour by absorbing heat is called boiling or vaporization. The temperature at which a liquid changes its state to gas is called boiling point. On reducing the temperature of the steam it becomes water again. The process by which a vapour is converted to liquid by releasing heat is called condensation. On reducing the temperature of water further to `0^@C`, it becomes ice which is a solid state of water.  The process by which a liquid is converted to solid by releasing heat is called freezing. The temperature by which a liquid changes its state to solid is called freezing point. Ice on heating, becomes water again by absorbing heat, a process known as MELTING . Dry ice changes directly to gaseous state without becoming liquid. This process is called SUBLIMATION. THUS, water changes its state when there is a change in temperature. 
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| 1823. |
A girl is wearing a pair of flat shoes. She weight 550 N. The area of contact of one shoe with the ground is 160 cm ^(2).What pressure will be exerted by the girl on the ground if she stands on two feet ? |
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| 1824. |
A cell is used to: |
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Answer» MEASURE CURRENT in a circuit |
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| 1825. |
If the displacement of an object is proportional to square of time, then the object moves with: |
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Answer» UNIFORM velocity |
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| 1827. |
Write the audible range of frequency for the normal human ear. |
| Answer» SOLUTION :20 HZ to 20 KHZ | |
| 1828. |
What are fluids ? Name two common fluids. (b) State Archimedes' principle. ( c) When does an object float or sink when placed on the surface of a liquid? |
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| 1829. |
A sphere of iron and another of wood the same radius are held under water. Compare the upthrust on the two spheres. |
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| 1830. |
What could be the finaltemperature ofa mixture of 100 g of water at 90^@C and 600 g of water at 20^@C |
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Answer» Solution :To FIND final temperature :`DeltaQ=me` 100 g of WATER originally at `90^@C` will loose an amount of heat , `DeltaQ=me DeltaT` `DeltaT=100 xx c xx (90-T)` The same amount of heat will be absorbed by 600 g of water originally at `20^@C` to raise its temperature to T . `DeltaQ`= 600 x c x (T-30) `600 C (T-20^@)=100C (90^@-T)` `6T-120^@=90^@-T` `6T+T=120^@+90^@` `7T= 210^@ rArr` T=210/7 `T=30^@C` |
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| 1831. |
Show how you prepare 1 L of 2(M) NaCl solution? |
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Answer» Solution :Calculation `M=(n)/(V("in L")),M=2=(n)/(1L)""therefore n=2mol=2xx58.5g" "NaCl=117.0g` NaCl So, you have to dissolve 117 g NaClin 1L water. Step 1 WEIGHT out 117 g NaCl. step 2: Transfer that AMOUNT of NaCl to a 1L (MARKED) FLASK. Step 3 : Add water and fill to 1L mark. |
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| 1832. |
A man standing with a load on his head or a man walking with load does no work scientifically. |
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Answer» Solution :Because a man standing with a load on his head EXERTS FORCE on the load in upward direction but that load has no displacement in the direction of force (in upward direction). Hence as per `W=Fs=Fxx0=0`, WORK done on the load is ZERO. So it can be said that aman standing with the load on his head does no work. Now, if the man with load is walking even though the man does no work on the load, because the man is EXERTING force on the load in upward direction, while as his motion is in horizontal direction, his displacement is in the horizontal direction. So, as the angle between the force and displacement is `90^(@)`, there is no work done on the load as per `W=F_(1)xxs=0xxs=0`. Where, `F_(1)` = Effective force along direction of dispalcement. |
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| 1833. |
The relative density of silver is 10.8. Find its density. |
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| 1834. |
The SI unit of electric current is kilogram . |
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| 1835. |
What is the value of relative density of water |
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| 1836. |
State which of the following situations are possible and give an example for each of these : (a) an object with a constant acceleration but with zero velocity. (b) an object moving in a certain direction with an acceleration in the perpendicular direction. |
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Answer» SOLUTION :(a) Yes. An object falling freely from a certain HEIGHT. (b) Yes. An object performing uniform CIRCULAR motion. [At every point on the circular path, the VELOCITY is tangential to the circle at that point and acceleration is radial towards the CENTRE of a circle.] |
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| 1837. |
Arrange the speed of sound in gases V_g, solids V_s and liquids V_l in an ascending order. |
| Answer» SOLUTION :`V_g LT V_l lt V_s` | |
| 1838. |
What is lactometer? Explain its principle and working. |
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Answer» Solution :One form of hydrometer is a lactometer, un instrument USED to check the purity of milk The lactometer works on the principle of gravity of milk. The lactometer CONSISTS of a long praduated test TUBE with a cylindrical bulb with the graduation ranging from 15 at the top to 45 at the bottom. The test tube is filled with air. This air chamber causes the instrument to float. The plencal bulb is filled with MERCURY to cause the lactometer to sink up to the proper level and to flout in an upright POSITION in the milk. Inside the lactoncler there may be a thermometer extendingfrom the bulb up into the upper part of the len tube where the scale is located. The correct lactometer reading is obtained only at the temperature of `60^(@) C`. A lactometer measures the cream content of milk More the cream, lower the lactometerfloats in the milk The average reading of normal milk is 32 The lactometers are used highly at milk processing units and at dairies. |
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| 1839. |
A body weighs 200 gf in air and 190 gf when completely immersed in water. Calculate: (i) the loss in weight of the body in water. (ii) the upthrust on the body. |
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Answer» Solution :Given: weight of the BODY in AIR =200 gf Weight of the body in water =190 gf (i) Loss in weight of the body =200gf-190gf =10gf (II) Upthrust on the body =loss in weight =10 gf. |
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| 1840. |
A solid weighs 30 gf in air and 26 gf when completely immersed in a liquid of relative density 0.8. Find : (i) the volume of solid, and (ii) the relative density of solid. |
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Answer» Solution :Given WEIGHT of solid in air `W_(1)=30` gf and weight of solid in liquid `W_(2)=26` gf. R.D. of liquid =0.8 `:.` DENSITY of liquid `=0.8gcm^(-3)` (i) LET V be the volume of solid. Weight of liquid displaced= Volume of liquid displaced `xx` density of liquid `xxg` `=Vxx0.8xxg` dyne `=Vxx0.8gf`.........i Loss in weight of the solid when immersed in liquid `=W_(1)-W_(2)=30-26=4gf` .........ii But the weight of liquid displaced in equal to the loss in weight of solid when immersed in liquid `:.` From eqns (i) and (ii) `Vxx0.8=4` or `V=4/90.8=5cm^(3)` (ii) Given weight of solid `=30gf` `:.` Mass of solid `=30g` Density of solid `=("Mass")/("Volume")=30/5=6gcm^(-3)` Hence relative density of solid =6 Alternative method (i) Volume of solid = Volume of liquid displaced = mass of liquid displaced / density of liquid `=(30-26)//0.8=5cm^(3)` (ii) R.D. of solid `("Weight of solid in air")/("Weight of solid in air"-"weight of solid in liquid")xx` R.D. of liquid `=30/(30-26)xx0.8=30/4xx0.8=6` |
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| 1841. |
Velocity (v) of sound in air , by vibrating resonating columns is found by "________" (l_(1) , l_(2) and n are first second resonating lengths and frequency of tuning fork used , respectively ) . |
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Answer» `V = 2(l_(2) - l_(1))` |
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| 1842. |
In hammerthrow, before the hammer is let go off, why is it whirled around along a circular path ? |
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| 1843. |
Find the free-fall acceleration of an object on the surface of the moon, if the radius of the moon and its mass are 1740 km and 7.4 xx 10^(22) kg respectively. Compare this value with free fall acceleration of a body on the surface of the earth. (AS_(1)) |
| Answer» SOLUTION :APPROXIMATELY `1.63 m//s^(2)` | |
| 1844. |
How can we obtain the restingpoint-indicating equilibrium, even with unequal masses in the two pans of a physical balance? |
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Answer» Solution :(i) CONSIDER the adjustments made to a PHYSICAL balance while determining the zero RESTING point. (ii) By adjustingscrews. |
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| 1845. |
The formula to find the linear speed of an object performing uniform circular motion is……… |
| Answer» SOLUTION :`V = (2PI R )/(t)` | |
| 1846. |
Take a toy car. Wind it using its key. Place the car on the ground. From where did it acquire energy? |
| Answer» Solution :The WORK DONE in WINDING is stored in it aspotential ENERGY. | |
| 1847. |
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. |
| Answer» Solution :Some leaves of a tree get detached when we shake its BRANCHES vigorously because branches come in MOTION while the leaves TEND to remain at rest due to INERTIA of rest. | |
| 1848. |
The spring balance is used to measure the mass of the given sphere. What is the mass of the sphere ? . |
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Answer» Solution :Mass of the solid = FINAL reading on the spring BALANCE - initial reading of the spring balance =`125 G - 10 g` =`115 g`. |
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| 1849. |
Phosphorescence is the emission of visible light when sunstances are irradiated with______. |
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| 1850. |
Engine of a car of mass 1500kg, keeps car moving with constant velocity 5ms^(-1). If frictional force is 1000 N, power of engine is ………….. |
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Answer» `5kW` `W=Fxxs` `THEREFORE P=(W)/(t)=(Fxxs)/(t)=Fxxv(because v=(s)/(t))` `thereforeP=(1000N)xx(5ms^(-1))=5000W` `P=5kW` |
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