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.
| 1901. |
The acceleration for electron and proton due to electrical force of their mutual attraction when theyare 1 A apart is: |
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| 1902. |
6. A satellite of mass m is orbiting the earth (of radiusR) at a height h from its surface. The total energyof the satellite in terms of go, the value ofacceleration due to gravity at the earth's surface,is[NEET (Phase-2) 2016mgoRmgoR2(1) ngA(2) 2(R+ h)(4) 2mgoR2R+ h2mgoR2 |
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| 1903. |
Find the height of the point vertically above theearth's surface at which the acceleration due togravity becomes 1% of its value at the surfacein terms of R. (R is the radius of the earth) |
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Answer» gh=gR^2/(R+h)^2 given gH=g/100 Therefore we have gR^2/(R+h)^2=g/100 or R+h=10R or h=9R |
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| 1904. |
2dI< Rcentre of earth is B, then its value at9. If acceleration due to gravity at distancefrom thedistance d above the surface of earth will be [whereR is radius of earth](1) BA2(2) 2d(R+d)3(R+ d)2(4) BA32 |
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Answer» answer is bR3/d(R+d)2i.e. if b is the acceleration at the centre of gravity is b.where R is the radius of earth. |
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| 1905. |
13. A solid sphere of mass m and radius r is placed insidea hollow thin spherical shell of mass M and radius R asshown in figure (11-E1). A particle of mass m' is placedon the line joining the two centres at a distance a fromthe point of contact of the sphere and the shell. Findthe magnitude of the resultant gravitational force on thisparticle due to the sphere and the shell if (a) r <x < 2r,(b) 2r <<2R and (c) x> 2R.and (e) x> |
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| 1906. |
1. Two velocities, each 5 ms^-1, are inclined to each other at 60°. Find their resultant. |
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| 1907. |
17. Two particles of masses "p" and "q" (p>q)are separated by a distance "d". The shiftin the centre of mass when the two particlesare interchanged is |
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| 1908. |
(5) Fill in the blank and rewrite the statement:Electric power -V^2/ |
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| 1909. |
1½Two particles of masses “p” and "g" (pq)are separated by a distance "d". The shiftin the centre of mass when the two particlesare interchanged is1) dop+q)/(p) 2 dp-q)/(prq)3) d p/(p-q66 -924) d q/ (p-q) |
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| 1910. |
Two particles start moving from same position on a circle of radius 20 cm with speed40π23.m/s and 36π m/s respectively in the same direction. Find the time after which theparticles will meet again. |
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Answer» Relative speed = 40π - 36π = 4π m/sec Circumference of the circle = 2π * 20 = 40π cm =0.4π m Time after which they again meet = 0.4π / 4π = 0.1 sec |
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| 1911. |
A pulley is attached to the ceiling of a lift moving upwards. Two particles are attached to the twoends of a massless string passing over the smooth pulley. The masses of the particles are in the ratio2 : 1. If the acceleration of the particles is g/2 w.r.t. lift, then the acceleration of the lift will be |
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Answer» Givenacceleration of the particles is g/2=a' now letacceleration of the lift = a upwards assume lift at rest and apply psudo force = mass* a to both masses ER. RAVI KUMAR ROY Let masses as 2m and m 2m>m so 2m will move downward with a' So 2mg+2ma-T=2ma'=2m(g/2)=mg .....(1) and similarly for m mass T-mg-ma=ma'=m(g/2) .....(2) Solving both we get a=(g/2) |
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| 1912. |
Two particles of equal mass have initial velocities 2i ms" and 2jms. First particle has a constantacceleration (i ms2 while the acceleration of the second particle is always zero. The centre ofmass of the two particles moves in(A) Circle(B) Parabola(C) Ellipse(D) Straight line |
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Answer» Vcm = m*2i +m*2j/(m+m) = i+j and Acm = m(i+j) +m*0/2m = i/2+j/2 so, Scm = Ut + 1/2*(acm)t² The is the equation of straight line motion option D. |
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| 1913. |
Two particles P and Q simultaneously startmoving from point A with velocities 15m/sand 20m/s respectively. The two particlesmove with acceleration equal in magnitudebut opposite in direction. When P overtakesQ at B its velocity is 30m/s. The velocity of Qat point B will beonemo |
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| 1914. |
57. A circular plate of uniform thickness has a diameter of 56 cm. Acircular portion of diameter 42 cm is removed from one edge of theplate. Find the centre of mass of the remaining portionAns. 9 cm from the centre of the plate. |
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Answer» Thickness of plate = t Totalmass of circularplate = MDensity of plate = M /(πr₁² t) , r₁ = 28 cm r₂= radius of cut out part= 21 cm Mass of small circular portion cut out = volume* density =πr₂² t *(M/π r₁² t) m₂ = M r₂² / r₁² Mass of the remaining part = m3 = M - Mr₂² / r₁² = M (r₁² - r₂²) / r₁² Let the center of mass of remaining portion be at a distance d from the center of original full plate. From symmetry the center of mass of remaining portion lies along the line joining the center of original plate and the center of removed part. Let Center of mass of total full plate = 0 0 = 1/M ( d * mass of remaining part + C₁C₂ * mass of removed part) 0 = d *M (r₁² - r₂²) / r₁² + 7 * M r₂² / r₁² 0 = d * (r₁² - r₂²) + 7 *r₂² d = - 7 * r₂² / (r₁² - r₂²) = - 7 * 21² / (28² - 21²) = - 7 *21 * 21/ 49*7 = - 9 cm Center gravity of the remaining portion is 9 cm to the left of the original center of full plate on the line of symmetry. Or, it is 19 cm from the edge of remaining plate along the line of symmetry. sristi patna se ho tum kon Institute me padhti ho |
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| 1915. |
14. A resistor is to be made of a material having aresistivity 4 x 10^-7 ohm metre. Its resistance is to be2 mili ohm and it is to be made of rectangularsection 5 mm × 2 mm. Calculate its length. |
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| 1916. |
200°C1. A resistance wire made from German silver has a resistance of 4.25 2. Calculateof another wire, made from the same material such that its 1m German silver has a resistance of 4.25 9. Calculate the resistanceof cross-section decreases by three times.rom the same material, such that its length increases by 4 times and area151 1bre |
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Answer» R = ρl/AR' = ρ(4l)/(A/3) = (ρl/A)(12) = 4.25 × 12 = 51 ohm |
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| 1917. |
14. A resistor is to be made of a material having aresistivity 4 x 10^-7 ohm metre. Its resistance is to be2 milli ohm and it is to be made of rectangularsection 5 mm x 2 mm. Calculate its length.[Ans. 0.05 m] |
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| 1918. |
4. One quarter section is cut from a unifomcircular disc of radius R. This section has amass M. It is made to rotate about a lineperpendicular to its plane and passingthrough the centre of the original disc. Itsmoment of inertia about the axis of rotation is(2001) |
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| 1919. |
If a body is to be projected verticallyupwards from earth's surface to reach aheight of 10 R, where R is the radius ofEarth, then the velocity required to do sc is241)gR21 20gR3)184) |
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| 1920. |
2. The work done to break a spherical drop of radius Rin n drops of equal size is proportional to(2) n1781/32/3(3) n1/3-1(4) nA/3 -1 |
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Answer» The work done to break a spherical drop of radius R in n drops of equal size is proportional to[1- 1/n⁰·³³] |
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| 1921. |
.What do you know about absorption and emission spectrums ? |
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Answer» Absorption spectrum is an electromagnetic spectrum in which a decrease in intensity of radiation at specific wavelengths or ranges of wavelengths characteristic of an absorbing substance (as chlorophyll) is manifested especially as a pattern of dark lines or bands. The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an atom or molecule making a transition from a high energy state to a lower energy state. |
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| 1922. |
2018the line ABor Way TUL?A point charge +10 PC is a distance 5 cm directly above the centreof a square of side 10 cm, as shown in Fig. 1.34. What is themagnitude of the electric flux through the square? (Hint: Think ofthe square as one face of a cube with edge 10 cm.)5 cm0° C isne lest de= -2.53m), resteat of theses alignedde 5 x 10dipoleave a me10 cmom10 cmtheiral |
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| 1923. |
In which three categories can plants be classified? |
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Answer» Herbs, shrubs and trees. |
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| 1924. |
सन; ने पक 80 (वन की9 Neitlo 'n,’;wc?\.r |
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Answer» Light travels faster in air than in glass because air is lighter medium compared to glass which is dense.i.e.glass partially resists the propagation of light through it (A) wave theory is correct option |
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| 1925. |
value tat wcd i |
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Answer» plz tell me if it's correct or not thanks a lot Katana........ |
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| 1926. |
stant force of 50 N is applied on a stationary objectof mass 10 kg How much distance would this object travelin ?s ? |
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| 1927. |
In a particular system, the unit of length, massand time are chosen to be 10 cm, 10 g and 0.1 J srespectively. The unit of force in this systemwill be equivalent to(1994) |
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| 1928. |
(18)The density of a material in CGS system ofunits is 4g cm. In a system of units in whichunit of length is 10cm and unit of mass is100 g, the value of density of material willbe-(a)0.04(c)40(b) 0.4(d) 400 |
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Answer» Density in CGS system is = 4 gms/ cm³Unit of length = 1 cm unit of mass = 1 gm Another system of units: unit of mass = gm' = 100 gm unit of length = cm' = 10 cm => 1 gm = and 1 cm = Substitute these values in the value of the density. Now density of the material: Thus in the new system the value of density will be 40 units. |
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| 1929. |
In FPS system which unit is the absolute unit ofweight?A NewtonB 1 PoundC 1 dyneD GramSem Jan-2015/WR) |
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Answer» In FPS system absolute unit of weight is Pound (B) is correct option b option is correct 1 pound |
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| 1930. |
a non metal with metallic lustrous |
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Answer» Graphite isthe non metalwhich hasmetallic lustre... -->Non metalsare mostly found in nature as Solids, Liquids and gases. --->Non metalsdo not posses themetallic lusterbecause they cannot reflect light due to the absence of free electrons in their structure. |
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| 1931. |
Explain the role of ablotie and blotic factors in soll formation. |
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| 1932. |
Section BGive reasons for the following:(a) Soil gets affected by continuous plantation of crops(b) Bread gets mould during rainy season.(c) Saving paper means saving trees.(d) Pure gold is not used for making jewellery(e) Dams are made broader at the base and narrower at the top. |
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Answer» a)Continuousplantation of crops in a field makes thesoilpoor in certain nutrients such as nitrogen, phosphorus, potassium, etc. Plants require nutrients for their proper growth and functioning b)Fungi thriveindark, moist and warm conditions, which is whybreadwill be become mouldy, more rapidlyduring the rainy season.During the rainy season, the air has a higher moisture content, which is ideal for fungal growth. c)Saving PaperisSavingthe Forests. ... In just a day, how many millions sheets ofpaperare used by Indonesian people, and thismeansthat there are millions of foresttreesare cut down to meet those needs. d)Pure goldis 24k and is extremely soft and malleable, thereby making it rare in jewellery.Goldof 20k, 18k and 14k are usuallyusedto makejewellery. The karat value ofgoldis reduced by mixing it with other metals, usually copper or silver. ... However, the value of thegoldalso decreases along with its karatage. e) This is because the pressure of the water is much greater deeper down and the damneeds to be thick at thebottom so that it is strong enough to withstand this larger pressure. |
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| 1933. |
8 Calculate the area enclosed by a urcle of diameter 108mto correct number of significan't figure. |
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| 1934. |
The range of masses we study in Physics is(a) 1027 kg to 1060 kg (b) 10-27 kg to 10s5 kg(c) 10-3° kg to 1055 kg (d) 10-30 kg to 1060 kg4. |
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Answer» option (c) is correct. |
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| 1935. |
g 10 m/s .8. Find the accelerations a,, a,, a, of the three blocksshown in figure (6-E8) if a horizontal force of 10 N isapplied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block.Take g = 10 m/s 22 kg3 kg7 kgH1 0.2a2азμ3= 0.0Figure 6-E8 |
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| 1936. |
The specific heat of a substance is0.09 cal/gm°C. If the temperature is measuredon Fahrenheit scale the value of its specificheat in cal/gm/°F is |
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| 1937. |
Example 2: Density of oil is 0.8 gm/cm. Find its value in MKSsystem. |
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Answer» density in MKS Will 800 kg/m×m×m 800 kg m^-3 is the density in MKS units. |
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| 1938. |
l EUlu? It is easter to pull a lawn roller then to push it explain.11. The velocity of a particle is v 4t' 2t-3. Find the displacement and acceleration for the object at -2 sec.Where v is in m/s. |
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| 1939. |
9.If is the angle between unit vectors A andis equal to(1 - AB)B, then(1 + A.B)1) tan? (/2)3) cot (012)2) sin(/2)4) cos? (012) |
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Answer» OP tion A hsjskslalskdkskfkri bshjdisk bdjdiufufrhr grhriririi hrjroro |
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| 1940. |
\sqrt { \left( \frac { 20 } { 18 \cdot 012 } \right) ^ { 2 } - 1 } |
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| 1941. |
A uniform wire of resistance 10 Ω is cut into three equal parts. These parts are now connected inparallel. Then the equivalent resistance of the combination is |
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| 1942. |
Define half life period. A radioactive isotope has a half life of 5 years. After how muchtime is its activity reduced to 3.125% of its original activity. S03 |
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| 1943. |
18.Half-life period oflead is:(A) 1590 years (B) 1590 days(C) Infinite(D) Zero |
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Answer» In total, 43 lead isotopes have been synthesized, with mass numbers 178–220. Lead-205 is the most stable radioisotope, with a half-life of around 1.5×10^7years. The second-most stable is lead-202, which has a half-life of about53,000 years, longer than any of the natural trace radioisotopes Lead 210 has a relatively short half-life of22.3 years. The amount of210Pb remaining in most samples is statistically insignificant after about seven half-lives. but this is not in option !! |
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| 1944. |
Two particles are moving in circular paths of radir, and r2 with same angular speeds. Then the ratioof their centripetal acceleration is46.(2) r, :互thle(3) r2 : r,(4)小時 |
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Answer» The formula for centripetal force is F = mv²/R sub. v =rw(omega) F =m r w² w (angular velocity) will be same as the time period is same...... F1 = m R1 W² F2 = m R2W² W(angular velocity) and mass are constant. F1/F2 = R1/R2. :- The ratio is R1 :R2. nhiaya |
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| 1945. |
-Calculatetheequivalentresistan202ANAMMMM22 |
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Answer» 3/2 omega is correct answer. |
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| 1946. |
httobdsinbotoeerruam ld th eth,3nrr he is t .ootis ed |
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Answer» option D is correct. D has the more.. curve surface.. so its focus will be nearest, among all the mirrors. |
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| 1947. |
1. A piece of wire of resistance R is cut into five equal parts. These parts are thenconnected in paraliel. If the equivalent resistance of this combination isR, then theratio R/RisIS(0l 25 |
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| 1948. |
A piece of wire of resistance R is cut into five equal parts. These parts are thenconnected in parallel. If the equivalent resistance of this combination is R then theratio R/R'is(a) 1/25(c) 5(d) 25 |
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| 1949. |
ece of wire of resistance R is cut into five equal parts. These parts are thenA piece ed in parallel. If the equivalent resistance of this combination is R,then theratio R/R'isal 1/25(c) 5(d) 25 |
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Answer» thank you for your reply |
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| 1950. |
A piece of wire of resistance R is cut into five equalparts. These parts are then connected in parallel.If the equivalent resistance of this combination isR', then the ratio R/R is(a) 1/251.(b) 1/5(c) 525CBSE 2012) |
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Answer» when the piece of wire having resistance R is cut into 5 . then the resistance of each parts becomes R/5. due to the change in length from l to l/5. now if the resistances are connected in parallel. the the resulting R' is 1/R' = 1/(R/5)+1/(R/5)+1/(R/5)+1/(R/5)+1/(R/5) => 1/R' = 5*(5/R) = 25/R => R' = R/25 so, the value of R/R' = 1/(1/25) = 25. |
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