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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.
| 97351. |
Suppose our government plans to build a Highway in our state.a. Which technique in geography can be used for its planning?. b. Which technique can be used to assess the extent of land lost? c. Which technique can be used to assess the compensation and to estimate the land holdings of similar size? |
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Answer» a. Geographic Information System b. Buffer analysis c. Overlay analysis |
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| 97352. |
Suppose you are analysing the satellite imageries of two strategically important locations A and B. The spatial resolutions of the imageries are 1km x 7km and 1m x 1m respectively.i. Which of these imageries are of high spatial resolution? ii. Which of these imageries cannot be utilized for microlevel studies?iii. What do you mean by the term spatial resolution? |
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Answer» i. Imagery B has high, spatial resolution. ii. Imagery A cannot be utilized for micro-level studies. iii. The size of the smallest object on earth that can be recognized by the sensor is the spatial resolution of that sensor. |
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| 97353. |
As a result of the landslide in Kavalappara in Malappuram district, a large area was destroyed and many became homeless. The Government decided to estimate the extent of land area destructed and to identify the real land owners using GIS. Which analytical capabilities of GIS can be made use of for each purpose? Explain. |
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Answer» To estimate the area of land destructed, buffer analysis can be used. The spatial data of the area of landslide are subjected to buffer analysis, and a special circular zone is created. This helps to estimate the houses and agricultural lands of the area. To find the real landowners, overlay analysis can be used. Overlay analysis is used for under-standing the mutual relationship among the various features on the earth’s surface and the periodic changes undergone by them. |
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| 97354. |
In each aerial photograph, nearly 60% of the places depicted in the adjacent photo is included.i. What is the significance in doing so? ii. Mention the name by which it is known. |
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Answer» i. For the continuity of photographs. To obtain three-dimensional view ii. Overlap |
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| 97355. |
Describe any three Formulas of Liquidity Ratio? |
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Answer» Liquidity refers to the ability of a firm to meet its financial obligations in the short-term which is less than a year. Certain ratios, which indicate the liquidity of a firm, are (i) Current Ratio, (ii) Acid Test Ratio, (iii) Turnover Ratios. It is based upon the relationship between current assets and current liabilities. i. Current Ratio = \(\cfrac{Current \,Assest}{Current \,liabilities}\) The current ratio measures the ability of the firm to meet its current liabilities from the current assets. Higher the current ratio, greater the short-term solvency . ii. Acid-test Ratio = \(\cfrac{Quick \,Assest}{Current \,liabilities}\) Quick assets are defined as current assets excluding inventories and prepaid expenses. The acid-test ratio is a measurement of firm's ability to convert its current assets quickly into cash in order to meet its current liabilities. iii. Inventory Turnover Ratio = \(\cfrac{Cost \,of\,Goods\,Sold}{Average \,Inventory}\) Turnover ratios measure how quickly certain current assets are converted into cash or how efficiently the assets are employed by a firm. |
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| 97356. |
Identify the major difference between Satellite images and Aerial photoA. Resolution B. Altitude C. Clarity D. Processing |
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Answer» Correct option: B. Altitude |
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| 97357. |
Enlist the steps you will undertake to create a new folder. |
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Answer» a) Double click the computer icon b) Select the drive in which you want to create a new folder c) Window will open up showing files & folders click new folder on the tool bar |
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| 97358. |
Which of the following is not a point of 7Cs of effective communication? (a) Conscious (b) Clear (c) Courteous (d) Complete |
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Answer» (a) Conscious Conscious is not a point of 7Cs of effective communication. |
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| 97359. |
Name different components of the starting system. |
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Answer» Starting system mainly consists of following parts: a. Battery b. Starter switch c. Starter motor d. Starter drive e. Heavy insulted cables from battery to starter motor f. Ignition switch |
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| 97360. |
what do you mean by bases ? give some example chemistry |
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Answer» those substance that are better in test Soaphy to touch and turn red litmus to Blue is called bases. example:- sodium hydroxide and chemical Hydroxide,etc. |
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| 97361. |
snapshot- Ranga 's marriage |
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Answer» As the narrator of the short story,Shyama, or as he calls himself 'the humble servant' of his readers,is instrumental in determining not just the course of the story but also the lives of the characters around him. He plays the dual role of creating and narrating the story, not only on paper but also in life as we seem him facilitating the marriage of Ranga and Ratna. The couple name their child after Shyama as a token of their appreciation for bringing them together. Shyama takes the reins of the story into his own hands and as a narrator,he maintains a level of camaraderie that is tinged with delightful candour to relate to us the story of Ranga's marriage. |
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| 97362. |
How is a shehnai different from a pungi? |
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Answer» The difference between a Pungi and Shehnai can be noticed by the difference in their shapes and the sound produced by them. A Shehnai is longer than a Pungi. The latter has a shrill, unpleasant sound whereas, the former has a soft, melodious sound. |
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| 97363. |
Why do Jim and Hans think that games or sports are good ways of resolving conflicts? Do you agree? |
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Answer» Jim and Hans thought that games or sports were the best ways to establish peace because it create a sense of coordination, discipline, fellow-feeling. I agree to this point. In games or sports no one dies nor does it make among orphan or widow. |
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| 97364. |
The length, breadth and thickness of a sheet are `4.234`m, `1.005` m and `2.01`cm respectively. Give the volume of the sheet to the correct significant figures. |
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Answer» Area of cubord `=2(lb+bh+hl)=2(4.234xx1.005+1.0005xx0.0201+0.0201xx4.234)=8.72 m^(2)` `"Volume of cuboid"=lxxbxxm=4.234xx1.005xx0.0201=0.0855 m^(3)` |
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| 97365. |
The currect voltage relation of diode is given by `1=(e^(1000 V//T)-1) mA`, where the applied voltage V is in volt and the temperature T is in degree Kelvin. If a student makes an error measuring `+-0.01 V` while measuring the current of `5 mA` at `300 K`, what will be error in the value of current in mA?A. `0.5 mA`B. `0.05 mA`C. `0.2 mA`D. `0.02 mA` |
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Answer» Correct Answer - C Given current `I=(e^(1000 V//T)-1) mA` `rArr 1+1=e^(1000V//T)` `dI=1000/T[e^((1000V)/T)]dV` `dI=1000/T[I+1]dV` `1000/300 [6]xx(0.01)` `dI=0.2 mA` |
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| 97366. |
Consider the meter bridge circuit without neglecting and corrections (`alpha`, `beta`) If the unknown Resistance calculated without using the end corraction, is `R_(1)` and using the end corrections is `R_(2)` thenA. `R_(1)gtR_(2)` when balanced point is in first halfB. `R_(1) lt R_(2)` when balanced point is in first halfC. `R_(1)gtR_(2)` when balanced point is in second halfD. `R_(1)gtR_(2)` always |
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Answer» Correct Answer - A If balance point is firsthalf say `l=40` If balance point is second half say `l=70` maximum permissible error in `p`: the specific resistivity of wire, from meter bridge is `p=(piD^(2)S)/(4L)(l)/(100-l)` Assume that know resistance in `RB(S)`, and total length of wire is presicely known, then lets find maximum permissible error in `p` due to error in `p` due to error in measurement of `l` (Balance length and `D`(diameter of wire). In `p=In((piS)/(4L)+2InD+Inl-In(100-l))` `(dp)/(p)=2(dD)/(D)+(dl)/(l)-(d(100-l))/((100-l))` `=2(dD)/(D)+(dl)/(l)-(dl)/(100-l)` `((dp)/(p))_(max)=2(DeltaD)/(D)+(Deltal)/(l)-(Deltal)/(100-l)` `((dp)/(p))_(max)` due to error in `l` only is `=(Deltal)/(l)+(Deltal)/(100-l)=(Deltal(100))/(l(100-l))` `((dp)/(p))_(max)` will be least when `l(100-l)` if max imum, i.e. `l=50cm` |
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| 97367. |
Calculate `(1001)^(1//3)`. |
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Answer» We can write `1001` as `1001=1000(1+1/1000)`, so that we have `(1001)^(1//3)[1000(1+1/1000)]^(1//3)=10[1+1/1000]^(1//3)` `=10(1+0.001)^(1//3)=10(1+1/3xx0.001)` `=10.003333` |
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| 97368. |
If the current changes from 5A to 3A in x sec and inductance is 10H. The emf is 10V, calculate the value of x.(a) 2s(b) 3s(c) 4s(d) 5s |
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Answer» The correct choice is (a) 2s To elaborate: We know that: emf=L(i2-i1)/t Substituting the values from the question, we get x=2s. |
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| 97369. |
Mating of two closely related individuals within the same breed is called _______. (A) in-breeding (B) out-breeding (C) out-crossing (D) cross-breeding |
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Answer» (A) in-breeding |
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| 97370. |
Calculate the resistance in an inductive circuit whose time constant is 2s and the inductance is 10H.(a) 7ohm(b) 10ohm(c) 2ohm(d) 5ohm |
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Answer» Correct option is (d) 5ohm To explain I would say: We know that: Time constant = L/R Substituting the values from the given question, we get R=5ohm. |
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| 97371. |
In AIDS, Kaposis sarcoma may respond to a. Interleukin – 2 infusion b. Azathioprine c. Alpha interferon d. None of these |
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Answer» In AIDS, Kaposis sarcoma may respond to Alpha interferon. |
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| 97372. |
Two bodies `A` and `B` of masses `10kg` and `5kg` are placed very slightly separated as shown in figure. The coefficient of friction between the floor and the bloks is `mu=0.4` . Block `A` is pushed by an external force `F`. The value of `F` can be changed. When the welding between block `A` and ground breaks, block `A` will start pressing block `B` and when welding of `B` also breaks, block `B` will start pressing the vertical wall - What should be the minimum value of `F`, so that block `B` can press the vertical wallA. `20N`B. `40N`C. `60N`D. `80N` |
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Answer» Correct Answer - C `F_(min)=f_(A)+f_(B)=60N`. |
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| 97373. |
A positively charged thin metal ring of radius R is fixed in the xy plane with its centre at the origin O. A negatively charged particle P is released from rest at the point `(0, 0, z_0)` where `z_0gt0`. Then the motion of P isA. periodic, for all values of `z_0` satisfying `0ltz_0ltoo`.B. simple harmonic, for all values of `z_0` satisfying `0ltz_0leR`C. approximately simple harmonic, provided `z_0lt lt R`D. such that P crosses O and continues to move along the negative z-axis toward `z = -oo`. |
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Answer» let Q be the charge on the ring the negative `-q` is released from point `P(0,0,Z_(0))` the electric field at P due to the chaged ring will be along positive z-axis and its magnitude will e `E=(1)/(4piepsi)(QZ_(0))/((R^(2)+Z_(0)^(2)-4pt)^(3//2))` therefore force on charge P will be toward the centre shown, and its magnitude is `F_(e)=qE=(1)/(4piepsi_(0))(Qq)/((R^(2)+Z_(0)^(2))^(3//2))Z_(0)` ..(i) similarly when it crosses the origin the force is again toward center O. thus the motion of the particle is periodic for all values of `Z_(0)` lying between 0 and `infty` secondly if `Z_(0) lt lt R,(R^(2)+Z_(0)^(2))^(3//2)toR^(3)` `F_(e)=(1)/(4pepsi_(0))xx(Qq)/(R^(3))xxZ_(0)` [from eq (i)] that is the restoring force `F_(e)prop-Z_(0)`, thence the motion of the particle will e simple harmonic (here negative sign implies that the force is towards its means position). |
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| 97374. |
A spherical symmetric charge system is centered at origin. Given, Electric potential `V=(Q)/(4piepsilon_0R_0)(rleR_0)`, `V=(Q)/(4piepsilon_0r)(rgtR_0)` A. For spherical region `rleR_0`. Total electrostatic energy stored is zero.B. Within `r = 2R_0`, total charge is Q.C. There will be no charge anywhere except at ` r = R_0` .D. Electric field is discontinuous at `r = R_0` . |
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Answer» For `rgtR_(0),E=(dphi)/(dr)=(Q)/(4piepsi_(0)r^(2))` therefore, charge enclosed by concentric spherical surface at r is `2R_(0)=epsi_(0)phi_(E)4pir^(2)=epsi_(0)(Q)/(4piepsi_(0)r^(2))=Q` For `rltR_(0),E=-(dV)/(dr)=0` and for `rgtR_(0)E=-(dV)/(dr)=4piepsi_(0)r^(2)` (here `V=phi`) |
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| 97375. |
A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric filed due to the sphere at a distance r from its centreA. increases as r increases, for rltRB. decreases as r increases, for `0 ltrltoo`C. decreases as r increases, for `Rltrltoo`D. is discontinuous at `r = R`. |
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Answer» The expressions of the electrical field are inside the sphere `(rltR),E=(1)/(4piepsi_(0))(Q)/(R^(3))r` Outside the sphere `(R lt r lt infty),E=(1)/(4piepsi_(0))(Q)/(r^(2))` Hence, E increases for `rltR` and decreases for `Eltrltinfty`. |
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| 97376. |
The deceleration exerienced by a moving motor blat, after its engine is cut-off is given by `dv//dt=-kv^(3)`, where `k` is constant. If `v_(0)` is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time `t` after the cut-off is.A. `(1)/(2ktV_(0))`B. `((V_(0))/(1+2kt V_(0)^(2)))`C. `(V_(0))/(sqrt(1-2k t V_(0)^(2)))`D. `(V_(0))/(sqrt(1+2kt V_(0)^(2)))` |
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Answer» Correct Answer - D `(dV)/(dt)=-kV^(3)` `underset(v_(0))overset(v)int V^(-3)dV=-kunderset(0)overset(t)intdt` `[(V^(-2))/(-2)]_(v_(0))^(v)=-kt` `[(1)/(V^(2))]_(v_(0))^(v)=2kt` `(1)/(V^(2))-(1)/(V_(0)^(2))=2kt` `(1)/(V^(2))-2kt+(1)/(V_(0)^(2))=(2ktV_(0)^(2))/(V_(0)^(2))` `V=(V)/(sqrt((1-2ktV_(0)^(2)))` |
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| 97377. |
A dielectric slab of thickness `d` is inserted in a parallel plate capacitor whose negative plate is at `x = 0` and positive plate is at `x = 3d`. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from 0 to `3d`:A. the magnitude of the electric field remains the sameB. the direction of the electric field remains the sameC. the electric potential increases continuouslyD. the electric potential increases at fist, then decreases and again increase. |
| Answer» Correct Answer - C | |
| 97378. |
In a second ODI match between England and India Bhuwnesh kumar bowled his medium fast bowling. He starts his run up from the distance of `50m` from the stumps (bowling end) and at the time of bowling a ball,his speed is `36(km)//(hr)` and the speed of ball is `144(km)//(hr)`. If his mass is `70 kg` and mass of the ball is `(1)/(4)kg` and the heat produce in his body is one tenth of the total work done by him, then Select the incorrect statement `:`A. Work done by the friction on him is negativeB. Work done by the ball on him is negativeC. Work done by ground on him is zeroD. Work done by the man on the ball is positive. |
| Answer» Correct Answer - A | |
| 97379. |
Find the reading of the ammeter when the voltmeter across the 3 ohm resistor in the circuit of Fig. reads 45 V. |
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Answer» Obviously I1 = 45/3 = 15A. If we take it as reference quantity, I1 = 3 ∠0º Now, Z1 = 3 −j3 = 4.24 ∠− 45º. Hence, V = I1Z1 = 15 ∠0º × 4.24 ∠−45º = 63.6 ∠−45º I2 = V/Z2 = (63.6 ∠ -45º)/(5 + j2) = (63.6 ∠-45º)/(5.4 ∠ 21.8º) = 11.77 ∠ − 66.8º = 4.64 − j10.8 I = I1 + I2 = 19.64 −j10.8 = 22.4 ∠28.8º |
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| 97380. |
Which among the following, is not a cause for power loss in a transformer- (i) Eddy currents are produced in the soft iron core of a transformer. (ii) Electric Flux sharing is not properly done in primary and secondary coils. (iii) Humming sound produed in the tranformers due to magnetostriction. (iv) Primary coil is made up of a very thick copper wire. |
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Answer» Answer is (iv) Primary coil is made up of a very thick copper wire. as Primary coil made of Thick Coper wire has very less R. Therefore negligible power loss. Rest all options are reasons for power losses in a transformer. |
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| 97381. |
At what distance (in metre) from the centre of the Moon,the intensity of gravitational field will be zero? (Take, mass of Earth and Moon as `5.98xx10^(24)` kg and `7.35xx10^(23)` kg respectively and the distance between Moon and Earth is `3085xx106(8)`m)A. zeroB. `3.85xx106(7)`C. `8xx10^(8)`D. `3.46xx10^(8)` |
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Answer» Correct Answer - B |
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| 97382. |
A vessel is partly filled with liquid. When the vessel is cooled to a lower temperature, the space in the vessel unoccupied by the liquid remains constant. Then the volume of the liquid `(V_L)` volume of the vessel `(V_V)` the coefficient of cubical expansion of the material of the vessel `(gamma_v)` and of the solid `(gamma_L)` are related asA. `gamma_(L) gt gamma_(v)`B. `gamma_(L) lt gamma_(v)`C. `(gamma_(v))/(gamma_(L))=V_(C)//V_(L)`D. `(gamma_(v))/(gamma_(L))=V_(L)//V_(C)` |
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Answer» Correct Answer - A `DeltaV_(L)=DeltaV_(V)rArr gamma_(L)V_(L)=gamma_(v)V_(v)` or `(gamma_(L))/(gamma_(v))=(V_(v))/(V_(L))` But `V_(v) gt V_(L)rArr gamma_(L) gt gamma_(v)` |
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| 97383. |
The emissive power of a black body at `T=300K` is `100Watt//m^(2)` consider a body B of area `A=10m^(2)` coefficient of reflectivity `r=0.3` and coefficient of transmission `t=0.5` its temperature is 300 K. then which of the followin is correct:A. The emissive power of B is `20W//m^(2)`B. the emissive power of B is `200W//m^(2)`C. the power emitted by B is `200watts`D. the emissivity of `B` is `=0.2` |
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Answer» Since, `e=a=0.2` (since `a=(1-r-t)=0.2` for the body B) `E=(100)(0.2)=20W//m^(2)` Power emitted `=e.A=20xx10=200` Watt |
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| 97384. |
An object follows a curved path. The following quantities may remain constant during the motionA. SpeedB. VelocityC. AccelerationD. Magnitude of acceleration |
| Answer» Correct Answer - A::C::D | |
| 97385. |
The value of acceleration due to gravity (a) is least on equator (b) is least on poles (c) is same on equator and poles (d) increases from pole to equator |
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Answer» The value of acceleration due to gravity is least on equator. |
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| 97386. |
The force required for producing tides in the ocean is. (a) 70% due to Moon and 30% due to Sun (b) 30% due to Moon and 70% due to Sun (c) 45% due to Moon and 55% due to Sun (d) 55% due to Moon and 45% due to Sun |
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Answer» (a) 70% due to Moon and 30% due to Sun |
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| 97387. |
The circuit has resistors of equal resistance R. The equivalent resistance between A and B, when key is closed is `R_1` and when key is open is `R_2`. Find the ratio `R_1 and R_2` . A. `11//12`B. `2//5`C. `1//5`D. none of these |
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Answer» Correct Answer - B b. A careful observation will reveal that one end of each resistor is connected to A and the other end of each resistor is connected to B. Hence, resistors are in parallel. `R_(eq) = R//5` When the key is open, current will pass through resistance 1 and 2 only, which are in parallel. `R_(eq) = R//2` |
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| 97388. |
In the following circuit, resistance of potentiometer wire is `100 Omega`. Power consumption in potentiometer wire is same when jockey is placed at 10 cm from end A or end B. Internal resistance of cell (r ) is A. `30 Omega`B. `6 Omega`C. `60 Omega`D. `3 Omega` |
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Answer» Correct Answer - A a. Case I : 10 cm from A `P_1= (epsilon/(r+10))^2 xx 10` Case II : 10 cm from B `P_2 = (epsilon/(r+90))^2 xx 90` `P_1 = P_2 implies r = 30 Omega` . |
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| 97389. |
A radiation of energy `E` falls normally on a perfctly refelecting surface . The momentum transferred to the surface isA. `E//c`B. `2E//c`C. `Ec`D. `E//c^(2)` |
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Answer» Correct Answer - B Initial momentum of surface `P_(i) = (E)/(C)` where c = velocity of light (constant). Since, the surface is perfectly reflecting so the same momentum will be reflected completely Final momentum `P_(i)=(E)/(C)`(negative value) `:.` Change in momentum `Delta_(p) = p_(f)-P_(i) = -(E)/(C)-(E)/(C) = -(2E)/(C)` Thus, momentum transferred to the surface is `Delta_(p)=|Delta_(p)|=(2E)/(C)`. |
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| 97390. |
STATEMENT-1: The activity of a radioactive sample decreases linearly with time. STATEMENT-2: The number of active nuclei present in a radioactive sample decreases exponentialy with time.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation, for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is a False, Statement-2 is True. |
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Answer» Correct Answer - D |
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| 97391. |
In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of first minima is(1) 25° (2) 20° (3) 30° (4) 45° |
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Answer» Answer is (1) For 2nd minima d sinθ = 2λ sinθ = √3/2 (given) ⇒ λ/d = √3/4 ...(i) So for 1st minima is d sinθ = λ sinθ = λ/d = √3/4 (from equation (i)) θ = 25.65° (from sin table) θ ≈ 25° |
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| 97392. |
List I(Types of precipitation)List II(Approximate size)A. MistI. 2 – 5mmB. SnowII. 0.5 – 5mmC. GraupelIII. 0.005 – 0.05mmD. SleetIV. 1mm – 20mmChoose the correct answer from the option given below:1. A - I, B - III, C - II, D - IV2. A - III, B - IV, C - II, D - I3. A - IV, B - II, C - III, D - I4. A - III, B - IV, C - I, D - II |
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Answer» Correct Answer - Option 4 : A - III, B - IV, C - I, D - II Correct matching:
The correct option is A-III, B - IV, C - I, D - II Precipitation:
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| 97393. |
A steel ball of `50mm` radius is kept on a hole of concrete. The diameter of the hole is `0.05mm` less then the diameter of the steel ball at `20^(@)C`. Coefficient of volume expansion of the steel ball is `3.2x 10^(-6)//^(@)C` . At what temperature the ball will be inside the hole ?A. `30^(@)C`B. `55^(@)C`C. `65^(@)C`D. `70^(@)C` |
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Answer» Correct Answer - D A steel ball ……….. `D_(s)-D_(H) = 0.05mm` Where `D_(s)` is diameter of ball and `D_(H)` is diameter of ball hole. As we increase temperature `D_(H)gtD_(s)` for ball to enter hole `D_(H)(1+12xx10^(-6)DeltaDeltaT)-D_(s)(1+(3.2)/(3)xx10^(-6)Delta T)gt0` `(12xx10^(-4)- (3.2)/(3)xx10^(-4))Delta Tgt0.05` Given `Delta T gt 46^(@)` |
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| 97394. |
A student working with a spring mass system finds the values of mass, spring constant and damping constant in `CGS` unit as `90, 4.5xx10^(+3)` and `180` respectively. What will be the time period of the spring mass system ?A. `2pi//7`B. `14pi`C. `pi`D. `2pi//sqrt(50)` |
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Answer» Correct Answer - A A student ………….. For damped oscillation `omega = sqrt((K)/(m) - (b_(1)^(2))/(4m^(2)))` `omega = sqrt((4500)/(90) - (180xx180)/(4xx90xx90)) = sqrt(50-1) = 7` Time period `= (2pi)/(7)sec` . |
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| 97395. |
A rod of length 1 m is released from rest as shown in the figure below.Ifω of rod is √n at the moment it hits the ground, then find n. |
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Answer» mg(l/2)sin30° = 1/2(ml2)/3ω2 Solving ω2 = 15 ω = √15 |
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| 97396. |
Two masses each with mass 0.10 kg are moving with velocities 3 m/s along x axis and 5 m/s along y-axis respectively. After an elastic collision one of the mass moves with a velocity 4i + 4j. The energy of other mass after collision is x/10 then x is. |
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Answer» KEi = KEf 1/2m x 25 + 1/2 x m x 9 = 1/2m x 32 + 1/2mv2 34 = 32 + v2 KE = 1/2 x 0.1 x 2 = 0.1 j = 1/10 x = 1 |
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| 97397. |
A ball of density ρ0 falls from rest from a point P onto the surface of a liquid of density ρ in time T. It enters the liquid, stops, moves up, and returns to P in a total time 3T. Neglect the viscosity, surface tension and splashing. The ratio ρ/ρ0 is equal to (a) 1.5 (b) 2 (c) 3 (d) 4 |
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Answer» Correct Answer is: (c) 3 The downward and upward motions must be symmetrical. Hence, the total time spent by the ball inside the liquid is T, i.e., the downward motion of the ball inside the liquid takes time T/2. Hence, its upward acceleration inside the liquid is 2g. Let its volume be V. Then, inside the liquid, Vρg - Vρ0g = Vρ0(2g) or ρ = 3ρ0. |
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| 97398. |
The resistance of ideal ammeter is-(A) Zero(B) very small(C) very large(D) infinite |
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Answer» Correct answer is (A) Zero |
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| 97399. |
When two bulbs of power 60 W and 40 W are connectred in series, then the power of their combination will be-(A) 100 W(B) 2400 W(C) 30 W(D) 24 W |
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Answer» Correct answer is (D) 24 W |
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| 97400. |
A 50-ohm resistor is in parallel with 100-ohm resistor. Current in 50-ohm resistor is 7.2 A. How will you add a third resistor and what will be its value of the line-current is to be its value if the line-current is to be 12.1 amp ?Read more on Sarthaks.com - https://www.sarthaks.com/457612/a-50-ohm-resistor-is-in-parallel-with-100-ohm-resistor-current-in-50-ohm-resistor-is-7-2-a |
| Answer» Source voltage = 50 × 7.2 = 360 V, Current through 100–ohm resistor = 3.6A Total current through these two resistors in parallel = 10.8 A For the total line current to be 12.1 A, third resistor must be connected in parallel, as the third branch, for carrying (12.1 − 10.8) = 1.3 A. If R is this resistor R = 360/1.3 = 277 ohms | |