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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.
| 1101. |
What is an ECG? How does it help in the field of medicine? |
| Answer» SOLUTION :The electrocardiogram (ECG) is one of the simplest and OLDEST cardiac investigations available. In ECG, the sound variations produced by heart is CONVERTED into electric signals. Thus an ECG is SIMPLY a representation of the electrical activity of the heart muscle as it changes with TIME. Usually it is printed on paper for easy analysis. The sum of this electrical activity, when amplified and recorded for just a few seconds is known as an ECG. | |
| 1102. |
The values of g at six,distance A,B,C,D and F form the surface of the earth are found to be 3.08 m//s^(2),9.23 m//s^(2),0.57 m//s^(2),7.34 m//s^(2),0.30 m//s^(2) and 1.49 m//s^(2) ,respectively. |
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Answer» Arrange these values of g according to the increasing distances from the SURFACE of the EARTH (keeping the value of g nearest to the surface of the earth first) |
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| 1103. |
Explain convection in daily life |
Answer» Solution :Convection in daily life: (i) Hot air BALLOONS: Air molecules at the bottom of the balloon get heated by a heat source and rise. As the warm air rises, cold air is pushed downward and it is also heated. When the hot air is trapped inside the balloon, it rises. (ii) Breeze During day time, the air in contact with the land becomes hot and rises. Now the cool air over the surface of the sea replaces it. It is called sea breeze. During night time, air above the sea is warmer. As the warmer air over the surface of the sea rises, cooler air above the land moves towards the sea . It is called land breeze.  (iii)Winds: Air FLOWS from AREA of high PRESSURE to area of low pressure. The warm air molecules over hot surface the and create low pressure. Do, cooler air with high pressure flows towards low pressure area. This causes wind flow (iv) Chimneys: Tall chimneys are kept in kitchen and INDUSTRIAL furnaces. As the but gases and smoke are lighter, they rise up in the atmosphere. |
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| 1104. |
A body of mass 10kg performs motion along a circle of radius 5m with speed of 10ms^(-1). Work done during one revolution is ………….J. |
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Answer» `2000pi` Where. `F_(1)` = Effective force along direction of displacement |
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| 1105. |
The heart of a man beats 75 times a minute. What isits (a) frequency ? |
| Answer» SOLUTION :`1.25 s^(-1)` | |
| 1107. |
The mass of an atom of oxygen is 16.0 u. Express it in kg. |
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Answer» Solution :Given:MASS of one ATOM of oxygen = 16.0 U 1 u = `1.66xx10^(-27) KG ` `therefore ` The mass of an atom of oxygen `= 16.0 xx1.66xx10^(-27) kg ` `= 2.656 xx 10^(-26) kg ` |
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| 1108. |
Water is cooled from 4^(@)C to 0^(@)C it: |
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Answer» CONTRACTS |
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| 1109. |
Twoplane mirrors X and Y are placed parallel to each other and are separated by a distance of 20 cm. An object is placedbetween the two mirrors at a distance5 cm from the mirror X. Find thedistance of thefirst three image formed in the mirror X. |
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Answer» Solution :(i) The OBJECT distance for thesecond IMAGE 'X'= a +d = B. Where 'd' is the distance between the two mirrors and a is thefirst image distance on Y. Thesecond image on Y is FORMED at a distance b + d where b is the first image distance on 'X'. The third image on X is formed at d + b + d. (ii) 5 CM, 35cm, 45cm |
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| 1110. |
A cylindrical vessel containing liquid is placed on the floor of an elevator. When the elevator is mad to acceleration equal to g, disscuss how the centre of gravity of the system containg vessel and liquid changes. Also, discuss how the centre of gravity of the system containing vessel and liquid changes when the elevator accelerates in the downward direction. Also discuss how the 'weight' would vary in each case. |
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Answer» SOLUTION :(i) Please consider the factors that effect the centre of gravity of a REGULAR and irregular bodies (ii) Consider the point as centre of gravity (G) of a body is AFFECTED, when elevator accelerates in UPWARDS direction and downward direction |
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| 1111. |
Why a truck or a motorbike has much wide tyres ? |
| Answer» Solution :The pressure exerted by it can be DISTRIBUTED to more AREA, and avoid the WEAR and TEAR of tyres. | |
| 1112. |
Temperature of a body is the measure of |
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Answer» sum total of kinetic and POTENTIAL energy of the molecules of the GIVEN body. |
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| 1113. |
Saurabh is a student of class IX whereas his younger brother Ashu studies in class VI. During the summer holidays, Saurabh and Ashu went to visit their uncle who lives in a village. Their uncle has a big mango orchard near the village which has produced a bumper crop of mangoes this year. On the way to village, Saurabh purchased a catapult (gulel) from a shop because he enjoys felling ripe mangoes from the mango trees of orchard with the help of catapult. On reaching his uncle's mango orchard, Saurabh gave a tiny piece of stone to Ashu and asked him to put it in the catapult and hit any mango on the tree. Ashu tried to throw away the stone with catapult without stretching the rubber strings of catapult. Due to this, the piece of stone fell down to the ground instead to reaching the mango on the tree. Saurabh then taught Ashu how to use the catapult properly. By using this catapult any tiny pieces of stones, Ashu was now able to fell many ripe mangoes from the orchard trees. Ashu then saw a beautiful bird sitting on the branch of a mango tree. When Ashu was aiming the catapult at the bird, Saurabh snatched the catapult from his hands quickly and scolded him. Meanwhile, Saurabh's uncle also reached to orchard. He was happy to see Saurabh and Ashu enjoying the mangoes which they had felled from the orchard trees. (a) Which types of energy is possessed by the stretched rubber strings of a catapult? (b) How do the catapult strings get this energy? (c) What energy transformation takes place when the stretched rubber strings of catapult throw away a piece of stone? (d) Why did the piece of stone just fall down when Ashu tried to throw it away without stretching the rubber strings of catapult? (e) When a mango attached to the trees is hit by a piece of stone thrown by catapult, the mango falls down. Which force causes mango to fall down? (f) Why didSaurabh prevent his brother Ashu from aiming catapult at the bird? (g) What values are displayed by Saurabh in this episode? |
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Answer» Solution :(a) The stretched rubber strings of a catapult posses 'elastic potential energy'. (b) When we do work in stretching the rubber strings of catapult, then the work done by us gets STORED in thestretched rubber strings in the form of elastic potential energy. (c) The stretched rubber strings possess elastic potential energy whereas the PIECE of stone thrown away by it possesses kinetic energy. So, the energy transformation taking place is: Elastic potential energy `rarr` Kinetic energy (d) The unstretched rubber strings of catapult do not possess elastic potential energy due to which the piece of stone is not thrown away, it just falls to the ground by the action of gravity. (e) The force of gravity causes MANGO to fall down. (f) Saurabh prevented his brother Ashu from aiming catapult at the bird because he did not want the bird to get injured or killed. (g) The values displayed by Saurabh in this episode are (i) Adventurous nature (because he went all the way to village to enjoy ORCHARD mangoes) (II) Helping nature (because he taught his younger brother the proper use to catapult), and (iii) Protection of wildlife (because he prevents his brother from injuring or killing the bird). |
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| 1114. |
When an object is at the centre of curvature of concave mirror the image formed will be virtual and erect. |
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Answer» Solution :FALSE. Correct Statement : The image formed is REAL, INVERTED and same SIZE of the object. |
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| 1115. |
The figure shows a jar containing honey and kerosene Which is the liquid labelled as X |
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| 1116. |
Name two quantities which you can calculate from speed-time graph. |
| Answer» Solution :Distance travelled by an object in GIVEN time interval and ACCELERATION PRODUCED in the object can be calculated from speed-time GRAPH. | |
| 1117. |
If an alpha-ray and a beta-ray have same kinetic energy, which of them has greater velocity and why ? |
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| 1118. |
An electric kettle of 2000 Wis used to boil 20 litres of water. Find the time required to boil to the water from its initial temperature of 20^(@)C. (Take specific heat of water as 4200 Jkg^(-1) K^(-1).Density of water = 1000kg m^(-3)) |
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Answer» Solution :(i) Power `P = E/t` Energysupplied by the electrickattle is used to increase the TEMPERATURE of 20 litres of water from `20^(@)C`to `100^(@)C`. But HEAT energy `Q = msDeltatheta`. Calculate the heat energy absorbed by the water. Substituting the value of energy and power in equation (1) we can determine the time. (ii) 56 minutes |
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| 1119. |
(a) A simple pendulum is made by suspending a bob of mass 500 g by a string of length 1 m . Calculate its time period at a place where g = 10 m s^(-2) . (b) How will the time period in part (a) the affected if bob of mass 100 g is used, keeping the length of string unchanged ? |
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Answer» SOLUTION :(a) Given : l=1 m , g = `10 ms^(-2)` Time period `T = 2pi sqrt(l/g) = 2 XX 3.14 xx sqrt(1/(10))` `=1.99 ` s (b) On CHANGING the bob by another bob of different MASS, the time period will remain unaffected because it does not depend on the mass of bob. |
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| 1120. |
A concave mirror produces three times magnified real image of an object placed at 7 cm in front of it. Where is the image located? (21 cm in front of the mirror). |
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Answer» Solution :GIVEN : `hi = 3h_o` `u= 7 CM` Solution: (i) `m = (h_i)/(h_o) RARR (3ho)/(h_o) therefore m =3`. (ii) `m = - (v)/(u)` ` v = -m xx u` `= -3 xx 7 cm = -21 cm`. HENCE, Real, inverted and magnified image will be formed at 21 cm in front of the mirror. |
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| 1121. |
A faulty mercury thermometer has a stem of uniform cross section marked in mm. If this reads 83 mm instead of 80 mm at LEP and 229 mm instead of 220 mm at UFP, find the difference in the length of the mercury thread in both the faulty and correct thermometers at 250^(@)C. (Take LFP =0^(@)C and UFP =100^(@)C) |
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Answer» Solution :(i) The relation comparing two SCALES is given by: `((S-LFP)/(UFP-LFP))_("correct")=((X-LFP)/(UFP-LFP))_("faulty")"(1)"` First find the length of the mercury thread at `250^(@)C` in correct thermometer by comparing it with celcius scale by using (1). Take it as `l_(1)`. Similarly, find the length of mercury thread at `250^(@)C` in faulty thermometer by comparing it with celcius scale. Make USE of (1) consider the length as `l_(2)." Find the VALUE "l_(2)-l_(1).` (ii) 18 mm |
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| 1122. |
Children under the age of 5 can hear upto. |
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Answer» 20kHz |
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| 1123. |
How does the speed of a giant wheel in an amusement park ? |
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| 1124. |
A body of volume 100cm^(3) weighs 1 kgf in air. Find (i) its weight in water and (ii) its relative density |
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| 1125. |
Kilometre is oneof the SI units of measurement . |
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| 1126. |
The ratio of displacement and distance travelled for a moving body is (always more than 1, always less than 1, less than 1 or equal to 1) |
| Answer» SOLUTION :LESS than 1 or EQUAL to 1 | |
| 1127. |
Amit buys few grams of gold at the poles as per instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not why [Hint : The value of g is greater at the poles than at the equator. |
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Answer» SOLUTION :Weight of a body on the earth is GIVEN by: W=mg. Where M= mass of the body G=Acceleration due to gravtiy The value of g is greater at POLES then at the equator. Therefore gold at the equator weighs les than at the poles. HENCE, Amit.s friend will not agree with the weight of the gold BOUGHT. |
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| 1128. |
A boy pushes a book kept on a table byapplying a force of 4.5 N. Find the work done by the force if the book is displaced through 30 cm along the direction of push. |
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Answer» SOLUTION :Force applied on the book (F) = 4.5 N Displacement (s) = 30 CM = (30/100) m = 0.3m Work done, `W = FS` ` = 4.5 XX 0.3` `1.35` J |
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| 1129. |
Each of the two guns,mounted on a rotating platform with their lengths parallel to each other, fires 20 bullets per second at a speed of 50 m s^(-1). If the perpendicular distance between the two guns is 1.2 m and the mass of each bullet is 25g, find the couple acting on the platform. |
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Answer» Solution :(i) FORCE = Impulse `div`Time (ii) Force on the GUN = force on the BULLETS = m `((v-U)/(t))`, where v and u are final and initial velocities of bullets, respectively, and m = mass of each bullet. (iii) Couple = force `xx` perpendicular distance between the forces. (iv) 30 N m |
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| 1130. |
What can you say about the nature of motion of a body if its displacement-time graph is (a) a straight line parallel to time axis ? (b) a straight line inclined to the time axis with an acute angle? (c) a straight line inclined to the time axis with an obtuse angle ? (d) a curve. |
| Answer» Solution :(a) BODY is STATIONARY (or no motion), (B) motion away from the starting POINT with uniform velocity (c) motion towards the starting point with uniform velocity (d) motion with variable velocity. | |
| 1131. |
A block of wood is floating on water with its dimensions 50 cmxx 50 cm xx50cm inside water. Calculate the buoyant force acting on the block. Take g=9.8Nkg^(-1) |
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| 1132. |
In a mercury barometer, if the tube containing mercury is tilted, then |
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Answer» height of MERCURY column remains same. |
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| 1133. |
How much heat energy is required to melt 5 kg of ice? (Specific latent heat of ice = 336 Jg^(-1)) |
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Answer» Solution :Given, m=5 kg =5000 g ,`L=336 Jg^(-1)` Heat energy REQUIRED = m x L =5000 x 336 =1680000 J or `1.68xx10^6` J |
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| 1134. |
Fielder giving a swing while catching a ball is an example of |
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Answer» Inertia of rest |
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| 1135. |
A particle moving with constant acceleration of 2m//s^(2) due west has an initial velocity of 9m/s due east. Find the distance covered in the fifth second of its motion. |
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Answer» Solution :Initial velocity U = +9 m/s Acceleration a = - 2 m/`s^(2)` In this problem, acceleration.s direction is opposite to the velocity.s direction. Let .t. be the time taken by the particle to reach a POINT where it makes a turn along the straight line. wehave , V = u + at O = 9 - 2 t We get , t = 4.5s Now let US find the distace convered in `(1)/(2)` second i.e. from 4.5 to 5 second Let u at t = 4.5 sec. Then distance convered in `(1)/(2) ` s. ` s= (1)/(2) at^(2)` s `= (1)/(2) xx 2 xx [ (1)/(2)]^(2)` `= (1)/(4)` m Total distance covered in fifth second of its motion is given by `S_(0) = 2s = 2 ((1)/(4)) = (1)/(2) `m . 
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| 1136. |
Convert the following . (i)35^@ C to Fahrenheit (.^@F) (ii)14^@F to .^@C |
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Answer» SOLUTION :(i)`T._(@F)=T._(@C)xx1.8+32` `T._(@F)=25^@Cxx1.8+32=77^@F` (ii)`T._(@C)=(T._(@F)-32)//1.8` `T._(@C)=(14._(@F)-32)//1.8=-10^@C` |
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| 1137. |
ldentify the relation between the first pair and complete the second. Hydrometer:Law of floatation. Excavator:_____. |
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| 1138. |
It is dangerous for loaded vehicles to negotiate a curve in the road without reducing speed. What is the reason ? |
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| 1139. |
When the earth revolves around the sun, the work done by the gravitational force of the sun is zero. |
| Answer» Solution :Because when the earth revolves around the sun, the GRAVITATIONAL force of the sun acting on the earth is TOWARDS the centre and displacement of the earth is TANGENTIAL to the CIRCULAR orbit. Thus, the displacement of the earth is perpendicular to the gravitational force of the sun. When displacement is perpendicular to the force, the work done `W=F_(1)xxs=0xxs=0`. So, when the earth revolves around the sun, the work done by the gravitational force of the sun is zero. | |
| 1140. |
Find the pressure at a depth of 10 m in water if tre atmospheric pressure is 100kPa. [1Pa=1N//m] [100kPa=10^(5) Pa=10 ^(5) N//m^(2) =1 atm.] (AS_(1)) |
| Answer» SOLUTION :198 KPA | |
| 1141. |
A microsocpe is provided with a main scale graduated with 20 divisions in 1 cm and a vernier scale with 50 divisions on it of length same as of 49 divisions of main scale. Find the least count of the microscope. |
| Answer» SOLUTION :0.001 CM | |
| 1142. |
A scooterist moves towards a vertical wall witha speed of 54 km h^(-1) . A person is standing on the ground and is behind the scooter , hears the sound . If the scooterist sounds the horn of frequency 400 Hz , calculate the apparent frequency sound heard by the person when (a) It is coming directly from the horn. (b) coming after reflection from the vertical wall. (Take speed of sound to be 330 m s^(-1)) |
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Answer» Solution :The speed of scooter = 54 km `h^(-1)` `= 54 xx 5//18= 15 m s^(-1)` The frequency of the sound of horn = 400 Hz (a) The observer with reference to the ground is at rest `implies` Velocity of observer = `0 m s^(-1)` The source of moving away from observer . `therefore` The velocity of source = `15 ms^(-1)` The apparent frequency heard by the observer is `= n^(1) = (V - V_(o))/(V + V_(s)) n = (330 - 0)/(330 + 15) xx 400` `= 0.956 xx 400 = 382 . 4` Hz (B) The apparent frequency of sound received after from the wall is `= n'' = ( 330)/(330 - 15) xx 400 = (330)/(315) xx 400 = 419` Hz [This is possible because the vertical wall reflects the sound without changing the frequency ]. |
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| 1143. |
What is law of conservation of momentum . |
| Answer» Solution :Momentum of TWO BODIES collision is EQUAL to the momentum after collision. In an isolated system, the TOTAL momentum remain conserved. | |
| 1144. |
Assertion: The accelerated motion of an object may be due to change in magnitude of velocity or direction or both of them. Reason: Acceleration can be produced only by change in magnitude of the velocity. It does not depend the direction. |
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Answer» If both assertion and reason are TRUE and reason is the CORRECT explanation of assertion |
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| 1146. |
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minute 20 s? |
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Answer» Solution :Diameter `= 200 m therefore r = 100 m` Time for one round of the track = 40 s `therefore` In 2 MIN 20 sec = 140 s, the athlete will complete `(140)/(40) = 3.5` rounds of the track. DISTANCE covered in one round = the circumference `=2pi r ` `therefore` distance covered in `3.5` rounds `= 2 XX (22)/(7) xx 100 xx 3.5= 2200 m` Displacement after `3.5` rounds, the athlete will be in the diametrically opposite position. `therefore ` displacement = diameter of the track = 200 m The distance covered `= 2200 m,` displacement `= 200 m.` |
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| 1147. |
Hetal having mass 50 kg climbs up 20 m height with an object of mass 30 kg for 40 s, then her power is …………… W. (Take g=10ms^(-2)) |
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| 1148. |
Ice is kept in double-walled container. Why? |
| Answer» Solution :Ice is kept in a double-walled container so as to prevent melting of ice from the HEAT ABSORBED present in the immediate surroundings. The VACUUM present in between the two walls prevents the transfer of heat from the first to the second WALL and hence the ice remains in the SOLID form for a longer time period. | |
| 1149. |
Guess which sound has a higher pitch : guitar or car horn ? |
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| 1150. |
What conclusion can be obtained from the observation that when the prongs of a sound making tuning fork touch the surface of water in a beaker, the water gets splashed? |
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