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.
| 8151. |
“Can I ask you something?” “Not now. _____ a moment.” A) At B) For C) On D) In |
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Answer» Correct option is D) In |
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| 8152. |
What prompted the young seagull to fly finally? |
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Answer» It was his hunger that prompted the young seagull to fly finally. He was very hungry and the mother would not give him the fish she had brought with her. She wanted him to fly to her to get it. |
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| 8153. |
What do you think is the real crisis faced by the young bird? |
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Answer» The real crisis faced by the young bird is his lack of confidence and his fear of going out of the protective shelter of his home. |
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| 8154. |
How did the family support the seagull? |
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Answer» The mother flew past him, making joyful sounds. His father flew over him, screaming with joy. His two brothers and sister were flying around him, soaring and diving. In this way the family supported him. |
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| 8155. |
Why was the young bird terrified? How did it overcome its fear? |
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Answer» The young bird dived at the fish his mother was carrying. When he dived he fell outwards and downwards into space. The wind rushed against his breast feathers and his stomach and wings. He could feel the tips of his wings cutting through the air. He was not falling how. He was soaring. He thus overcame his fear. |
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| 8156. |
“ Here’s a birthday present _____ you.” “Oh, thank you!” A) by B) in C) for D) at |
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Answer» Correct option is C) for |
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| 8157. |
In 200-250 words write a story based on the input given below : Two teams - in the playground - whistle blew - match about to begin - the two captains looked tense - suddenly there was a commotion. |
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Answer» It was a great opportunity for me as I got a chance to watch a football match between Mohan Baghan of India and Massay club of Sri Lanka. I had already taken my seat. Both the teams were on the ground. It was a beautiful play ground. There was greenery, around it and different kinds of flowers were blooming. The weather was very charming and a gentle, cold breeze was blowing. I was excited to watch the match. The audience was also excited. They were blowing whistles and shouting slogans. All of us had been waiting for the match to start. The referees were also on the ground. The whistle blew which was a signal to both the teams to line up as the Prime Minister was in the stadium for inauguration. After formal introduction, the match was about to begin. The players of both the teams were in full zeal to show their game. All of a sudden. there was an announcement to be beware of the unknown things and keep them away. There was a secret report of some terrorists who were there. The two captains looked tense. No body knew, what to do. Suddenly, there was a commotion in one of the corners of the stadium. The police was there and they were holding two persons by their collar and took them away. The two terrorists had been arrested. Now, there was another announcement that everything was allright. The warning had been taken back. Now bottle the captains along with the audience were pleased. There was a great thrill in the match. Mohan Baghan won the match by 3 to 1 goals. Everyone was delighted. |
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| 8158. |
Panthera jeo is the scientific name of Lion. List the rules you follow to write this scientific name. |
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Answer» 1. Biological names are generally in Latin and written in italics. 2. When handwritten, the words in a biological name are separately underlined. 3. The first word in a biological name represents the genus and second word represents the species name. 4. The first name (Genus) starts with capital letter and the second name (species) starts with small letter. |
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| 8159. |
Figure shows crosssection view of a infinite cylindrical wire with a cavity, current density is uniform `vec(j)=-j_(0)hat(k)` as shown in figure A. magnetic field inside cavity is uniformB. magnetic field inside cavity is along `vec(a)`C. magnetic field inside cavity is perpendicular to `vec(a)`D. If an electron is projected with velocity `v_(0)hat(j)` inside the cavity it will move undeviated. |
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Answer» Correct Answer - A::C::D Magnetic field is given by `vec(B) =(mu_(0)(vec(j)xxvec(a)))/2` |
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| 8160. |
The electric field in a region is given by `E = (E_0x)/lhati`. Find the charge contained inside a cubical 1 volume bounded by the surfaces `x=0, x=a, y=0, yh=, z=0` and z=a. Take `E_0=5xx10^3N//C` l=2cm and a=1m. |
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Answer» Correct Answer - `2.2xx10^(-12) C` `q=in_(0) int vec(E).dvec(S)=in_(0) int (E_(0)x)/l hat(i)` `(dxhat(i)+dyhat(j)+dzhat(k))=in_(0)xx[(E_(0)x^(2))/(2l)]_(0)^(a)` Put given values `rArr 2.2xx10^(-12) C` |
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| 8161. |
A small object is placed at distance of 3.6 cm from a magnifier of focal length 4.0 cm. a. find the position of theimage. B. find the linear magnification c. Find the angular magnification.A. 10B. 7C. 5D. None of these |
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Answer» Correct Answer - B If the object is placed at a distance `u_(0)` from the lens, the angel subtended by the object on the lens is `beta = (h)/(u_(0))` where h is the height of the object. The maximum angle subtended on the unaided eye is `alpha = (h)/(D)` Thus, the angular magnification is `m = (beta)/(alpha) = (D)/(u_(0)) = (25 cm)/(3.6 cm) = 7.0` |
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| 8162. |
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth.A. The acceleration of S is alwas directed towards the centre of the EarthB. The angular momentum of S about the centre of the Earth changes in direction, but its magnitude remain constantC. The total mechanical energy of S varies perodically with timeD. The linear momentum of S remains constant in magnitude |
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Answer» Correct Answer - A Force on satellite is always towards earth, therefore, acceleration of satellite S is always directed towards centre of the earth Net torque of this gravitational force F about centre of earth is zero. Therefore, angular momentum (both in magnitude and direction) of S about centre of earth is constant throughout. Since the force F is conservative in nature, therefore mechanical energy of satellite remains constant. speed of S is maximum when it is nearest to earth and minimum when it is farthest. |
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| 8163. |
What is mgnification |
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Answer» The ratio of the height of an image to the height of an object is called magnification. |
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| 8164. |
An object is seen thorugh a simple microscope of focal length 12 c. Find the angular mgnification produced if the image is formed at the near pointofhe eye which is 25 cm away from it.A. 37/12B. 25/12C. BothD. None |
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Answer» Correct Answer - A The angular magnification produced by a simple microscope when the image is formed at the near point of the eye is given by `m=1+(D)/(f)` Here `f =12 cm,D=25 cm` Hence `m-1 +(25)/(12) =3.08`. |
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| 8165. |
In which of the case, a man wearing glasses of focal length, \( +1 m \) cannot clearly see beyond \( 1 m \).(A) if he is farsighted(B) if he is nearsighted(C) if is vision is normal(D) in each of these cases |
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Answer» Answer: in each of these cases The man is wearing glasses of positive power (converging lens). Hence, he cannot see nearby objects clearly. In other words, he is farsighted. Since he cannot see beyond 1 m, he is nearsighted. If a person with normal vision wears glasses of focal length +1 m, then the person will not be able to see beyond 1 m. |
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| 8166. |
How many images can we see at a time busying two mirrors? |
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Answer» The number of images changes accordance with the relation between the angle between the mirrors. |
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| 8167. |
Angle(θ)Number of images (n)4560901201801.How many images can be seen when viewed from A and B?2. What if viewed from other positions in between the mirrors?3. How much is the angle between the mirrors?4. What is the relation between the angle between the mirrors and the number of images? |
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Answer»
1. 3 2. 3 3. 90° 4. Number of images = n = \(\cfrac{360}{\theta}-1\) |
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| 8168. |
An object is placed in front of a concave mirror 20 cm away from it. If its focal length is 40 cm, locate the position of image and its nature. |
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Answer» f = – 40 cm, u = – 20 cm f = \(\cfrac{uv}{u+v}\) – 40 = \(\cfrac{-20}{-20+u}\) – 40 ( – 20 + v ) = – 20 v – 20 + v = -20/-40 = v/2 \(\therefore\) - v/2 = -20 Features of image: At behind the mirror, very large, virtual, direct. |
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| 8169. |
An object is placed 8 cm away in front of a concave mirror of focal length 5 cm. Find out the position of image and magnification. Find out whether the image is inverted or erect by drawing the ray diagram on a graph paper. |
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Answer» f = – 5 cm, u = – 8 cm -5 = \(\cfrac{-8v}{-8+v}\) -5 (-8 +v) = -8v 40-5v = -8v 40 = -3v \(\therefore\) 3v = -40 v = -40/3 = -13.34 cm Features of images : Beyond C, big, real, inverted Magnification m = \(\cfrac{h_i}{h_o} = \cfrac{-v}{u}\) m = \(\cfrac{-v}{u}\) = (-13.34/-8) = -16.7 |
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| 8170. |
What is hydrotropism ? How would you demonstrate hydrotropism with the help of an experiment. |
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Answer» Hydrotropism : The response of a plant part to water is called Hydrotropism, If the plant part moves towards water, it is called hydropism. Experiment procedure : (i) Plant a seedling in a vessel containing soil. (ii) Adjacent to the seedling put a porous pot containing water. (iii) Leave the soil up for few days. Observation On examinig the roots of seedling it is observed that the roots bend towards the source of water and do not grow straight. Results : Root grows towards water called as hydrotropism. |
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| 8171. |
The given figure shows the image formation by a concave mirror. Analyse the figure and write down different measures using New Cartesian Sign Convention. |
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Answer»
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| 8172. |
Aren’t the focal length of \(\cfrac{uv}{u+v}\) the mirror and the average value of obtained from the table the same? |
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Answer» Yes, the focal length of \(\cfrac{uv}{u+v}\) the mirror and the average value of obtained from the table the same. |
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| 8173. |
Record the measurements shown in the figure using the New Cartesian Sign Convention.1. Distance to the object from the mirror (u) =2. Distance to the image from the mirror (v) =3. Height of object (OB) =4. Height of image (IM) = |
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Answer» 1. Negative. 2. Negative. 3. Positive. 4. Negative. |
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| 8174. |
Oxytocin and ADH are produced by hypothalamus and released from (1) Anterior pituitary (2) Posterior pituitary (3) Pineal gland (4) Thymus |
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Answer» (2) Posterior pituitary |
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| 8175. |
A dental doctor uses a mirror of focal length 8 cm. To see the teeth clearly what should be the maximum distance between the teeth and the mirror? Justify your answer. Which type of mirror has been used by the doctor? |
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Answer» The dental doctor uses a concave mirror. Effect and enlarged image can be obtained using the mirror. Such images are formed when the object is kept in between the main focus and pole of a concave mirror. So the minimum distance between the mirror and the teeth must be in 8 cm to view the teeth clearly. |
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| 8176. |
From the above table, find out which mirror always gives an erect and diminished image and write it down. |
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Answer» The image formed by a convex mirror is always erect and diminished. |
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| 8177. |
Why it is written on rearview mirrors that “Objects in the mirror are closer than they appear”. |
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Answer» The image formed by a convex mirror is always erect and diminished. Hence the driver who sees the image of vehicles on the mirror develops a feeling that the vehicles coming from behind are at a greater distance. This may turn out to be dangerous. |
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| 8178. |
An object is moving in a straight line. Its position versus time graph is shown below. There is a detector at origin which analyze the motion of object between `t=6` and `t=8` seconds, which statement describes the motion shown in the graph. A. detector observes the object to be moving towards origin.B. detector observes the object to be moving away from origin.C. Detector observes the motion at decreasing speed.D. Detector observes the motion at increasting speed. |
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Answer» Correct Answer - A From `t=6 sec` to `t=8 sec`. Velocity of partcle is towards `-ve` `x-` axis and position is positive so object is moving towards origin. |
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| 8179. |
A thermodynamic process obeys the following relation `2dQ=dU+2dW` where dQ, dU, dW has ussual meaning then at which statement is true [given di-atomic gas, R=gas constant] then heat capacity for the process isA. `(5R)/(2)`B. `(7R)/(2)`C. `(3R)/(5)`D. infinite. |
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Answer» Correct Answer - D `2dQ=dU+2dW` `dQ=dU+dW` `dQ=dW` `impliesdU=0` process is isothermal |
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| 8180. |
The pressure of an ideal gas varies according to the law `P = P_(0) - AV^(2)`, where `P_(0)` and `A` are positive constants. Find the highest temperature that can be attained by the gasA. `(2P_(0))/(3R)(P_(0)/(3alpha))^(1//2)`B. `(2P_(0))/(2R)(P_(0)/(3alpha))^(1//2)`C. `(P_(0))/(R)(P_(0)/(3alpha))^(1//2)`D. `(P_(0))/(R)(P_(0)/(alpha))^(1//2)` |
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Answer» Correct Answer - 1 `P=P_(0)-alphaV^(2)` `("nRT")/("V")=P_(0)-alphaV^(2)` `RT=P_(0)V-alphaV^(3)" "[becausen=1]` `(dT)/(dV)=(P_(0))/(R)-(3alphaV^(2))/(R)=0` `V=sqrt((P_(0))/(3alpha))` `RT=P_(0)sqrt((P_(0))/(3alpha))-alpha(P_(0))/(3alpha)sqrt((P_(0))/(3alpha))` `T_("max")=(sqrtP_(0))/(sqrt3alphaR)(P_(0)-(P_(0))/(3))=(2P_(0))/(3R)sqrt((P_(0))/(3alpha))` |
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| 8181. |
A soap bubble (surface tension`=T`) is charged to maximum surface density of charge`= sigma`. When it is just going to burst. Its radius `R` is given by:A. `R=(sigma^(2))/(8epsilon_(0)T)`B. `R=8epsilon_(0)T/(sigma^(2))`C. `R=(sigma)/(sqrt(8(epsilon_(0)T))`D. `R=(sqrt(8epsilon_(0)T))/(sigma)` |
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Answer» Correct Answer - B The pressure due to surface tension `=(4T)/R` The pressure due to electrostatic forces `=(sigma^(2))/(2epsilon_(0))` Just before the bubble bursts `(4T)/R=(sigma^(2))/(2epsilon_(0))` or `R=(8Tepsilon_(0))/(sigma^(2))` |
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| 8182. |
Suppose the earth suddenly shrinks in size, still remaining spherical and mass unchanged (All gravitational forces pass through the centre of the earth).A. The days will become shorter.B. The kinetic energy of rotation about its own axis will decreaseC. The duration of the year will increaseD. The magnitude of angular momentum about its axis will increase. |
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Answer» Correct Answer - A `I_(1)omega_(1)=I_(2)omega_(2)` (Angular momentum is conserved) As `I_(2)` decreases `omega_(2)` increases. Thus `T=(2pi)/(omega)` i.e. `T` decreases Therefore the earth is completing each circle around its own axis in lesser time. `K.E.=1/2Iomega^(2)` Therefore `K.E.` of rotation increases. Duration of the year is dependent upon time taken to complete one revolution around the sun. |
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| 8183. |
Virtual library is a library without _______. (a) Collection (b) Resources (c) Wall (d) Information |
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Answer» (c) Wall Virtual library is a library without Wall. wall is the correct answer. A Virtual Library is a collection of resources available on one or more computer systems, where a single interface or entry point to the collections is provided. |
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| 8184. |
Ihusitamon \( 15.3 \). Find the equation of the circle having radias 5 and which touches line \( 3 x+4 y-11=0 \) at point \( (1,2) \). Solution. |
Answer» let the equation of circle be-\(x ^2+y ^2+2gx+2fy+c=0\) centre of circle = (-g,-f) radius of circle = \( \sqrt{f^2+g^2-c } = 5(given)\) distance between the centre of circle and the line 3x+4y-11=0 is radius of the circle, as it touches the circle, i.e.5 therefore, \( {|-3g-4f-11|\over \sqrt{3^2+4^2} }=5\) → -3g-4f-11=25 → -3g-4f=36 ....(1) now slope of normal to the circle at (1,2) \(m = {2+f \over 1+g}\) slope of the given line, \(m' = {-3 \over 4}\) therefore mm'=-1(normal and tangent to the circle at a point are always perpendicular) →\( {2+f \over 1+g}={4\over 3}\) → 6+3f=4+4g → 4g-3f=2 ....(2) multiply equation (1) and (2) by 4 and 3 respectively, → -12g-16f=144 ....(3) → 12g-9f=6 ....(4) add equation (3) and (4) we get, → -25f=150 → f=\({150\over -25}=-6\) substituting value of f in equation (1). → -3g+24=36 → \(g= {36-24\over -3}=-4\) thus the centre of circle is (4,6) now, put the values of g and f in radius of circle equation \( \sqrt{f^2+g^2-c } = 5(given)\) → \( \sqrt{16+36-c } = 5\) sqaring both sides we get, 52-c=25 → c= 52-25= 27 hence the equation of circle is \(x^2+y^2-8x-12y+27=0\) |
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| 8185. |
what made gulliver scared and confounded |
| Answer» The Lilliputians is the answer. | |
| 8186. |
5. A stone is thrown vertically upward with an initial velocity of \( 40 m / s \). Taking \( g=10 m / s ^{2} \), find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? |
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Answer» 40m/s rbs a new to the law
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| 8187. |
Solve 2x2 – 5x + 2 = 0 |
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Answer» 2x²-5x+2=0 2x²-4x-x+2=0 (2x²-4x)-(x-2)=0 2x(x-2)-(x-2)=0 (x-2)(2x-1)=0 x-2=0. And. 2x-1=0 x=2. And. x=½Given quadratic equation is 2x²-5x+2=0 2x²-4x-x+2=0 (2x²-4x)-(x-2)=0 2x(x-2)-(x-2)=0 (x-2)(2x-1)=0 x-2=0. And. 2x-1=0 x=2. And. x=½ |
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| 8188. |
If `s= ut +(1)/(2) at^(2)`, where u and a are constants. Obtain the value of `(ds)/(dt)`. |
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Answer» Correct Answer - u + at `s= ut +(1)/(2) at ^(2) therefore (ds)/(dt) = u+ at` |
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| 8189. |
If `sec theta = (5)/(3) and 0 lt theta lt (pi)/(2)`. Find all the other T-ratios. |
| Answer» `sintheta=(4)/(5), cos theta=(3)/(5),tantheta=(4)/(3), cos theta=(3)/(4), cosec theta=(5)/(4)` | |
| 8190. |
Distance between two point `(8, -4)` and `(0, a)` is `10`. All the values are in the same unit of length. Find the positive value of a. |
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Answer» Correct Answer - 2 Let P(8, -4) and Q(0, a) be two points, distance PQ is 10. `therefore " "` According to distance formula PQ = `sqrt((x_2-x_1)^(2) + (y_2-y_1)^(2))` `" "=sqrt((0-8)^(2) + (a+4)^(2))` `rArr PQ= sqrt (64+ (a^(2) + 16 + 8a)) ` `rArr a^(2) + 8a + 80 = 100 rArr a^(2) + 8a =20` `rArr a^(2) + 10 a -2a -20=0` `rArr (a-2)(a+ 10)=0 rArr (a=2) and (a=-10)` `therfore " "a=2` |
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| 8191. |
Mr. Mahajan purchased 100 shares, each of face value Rs. 100, when the market price was Rs. 45 per share, paying 2% brokerage. If the rate of GST on the brokerage is 18%, find the total amount he spent. |
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Answer» Amount spent to purchase 100 shares = 45 × 100 = Rs 4500 Brokerage = 4500 × 2 / 100 = Rs 90 GST on brokerage = 90 × 18 / 100 =Rs 16.20 ∴ Total amount = 4500 + 90 + 16.20 = Rs 4606.20 |
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| 8192. |
There are 25 rows of seats in an auditorium. The first row is of 20 seats, the second of 22 seats, the third of 24 seats, and so on. How many chairs are there in the 21st row ? |
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Answer» The arrangement of chairs is 20, 22, 24, 26, .......... Which is an A. P. Here, a = 20, d = 2. We want to find t21 . tn = a + (n-1)d ∴ t21 = 20 + (21-1) × 2 = 20 + 40 = 60 ∴ There are 60 chairs in the 21st row. |
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| 8193. |
Find the first term and common difference for each of the A.P. 5,1,-3,-7,... |
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Answer» First term=5 |
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| 8194. |
The diameter of sphere is 42cm then its surface area is –(A) 1386 cm2 (B) 4158 cm2 (C) 5544 cm2 (D) 2772 cm2 |
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Answer» Correct answer is (C) 5544 cm2 |
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| 8195. |
What is the common difference of an A.P in which a18 – a14 = 32? (A) 4 (B) -4 (C) 8 (D) -8 |
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Answer» Correct answer is (C) 8 |
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| 8196. |
In a two digit number, the digit of the unit place is double the digit in the tens place. The number exceeds the sum of its digit by 18. Find the number. |
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Answer» Let the digit in the tens place of the number be x. |
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| 8197. |
The sum of two numbers is 15. Sum of its reciprocals is 3/10. Find the numbers. |
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Answer» Let x, 15 – x be the numbers \(\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}\) \(\frac{15-x+x}{15x-x^2}=\frac{3}{10}\) 150 = 45x - 3x2 3x2 - 45x + 150 = 0 x2 - 15x + 50 = 0 \(x=\frac{-15\pm \sqrt{225-200}}{2}=\frac{15\,\pm\,5}{2}=10,5\) Numbers = 10, 5 |
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| 8198. |
The roots of the quadratic equation ax2 + bx + c = 0 (a ≠ 0) will be reciprocal of each other, if (A) b = c (B) c = a (C) a = b (D) none of these |
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Answer» Correct answer is (B) c = a |
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| 8199. |
What is the HCF of smallest prime number and the smallest composite number? |
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Answer» The required numbers are 2 and 4. HCF of 2 and 4 is 2. |
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| 8200. |
If one root of the quadratic equation 3x2 – 10x + k = 0 is reciprocal of the other. Then find the value of k. |
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Answer» Given that one root of equation 3x2 − 10x + k = 0 is reciprocal to other root. Hence, let the roots are α and \(\frac{1}{\alpha}.\) Therefore, product of roots = α . \(\frac{1}{\alpha}=\frac{k}{3}.\) \(\frac{k}{3}=1 \Rightarrow k = 3.\) Hence, the value of k is 3. |
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