HBSE Class 11 Sample Paper 2026 Physics

BSEH SAMPLE PAPER (2025-2026)
Class 11th (Sr. Secondary)
Roll No
PHYSICS
(Hindi and English Medium)
ACADEMIC/OPEN
[Time allowed: 3 hours] [Maximum Marks: 70]

lkekU; funsZ'k%
1. lHkh iz'u vfuok;Z gSA
2. iz'u i=k esa oqQy 35 iz'u gSA
3. bl iz'u&i=k ik¡p [k.M gSaA [k.M&v, [k.M&c, [k.M&l, [k.M&n, vkSj [.aM&;A ;s lHkh [k.M vfuok;Z gSaA
4. [k.M&v esa 1 vad osQ 18 oLrqfu"B iz'u gSaA
5. [k.M&c esa 2 vad osQ 7 vfry?kq mÙkjh; iz'u gSaA
6. [k.M&l esa 3 vad osQ 5 y?kq mÙkjh; iz'u gSaA
7. [k.M&n esa 5 vad osQ 3 nh?kZ mÙkjh; iz'u gSaA
8. [k.M&; esa 2 osQl LVMh vk/kfjr iz'u gSa] izR;sd 4 vad osQ gSaA
9. dksbZ lexz fodYi ugha gSA ;|fi [k.Mµc] l] n vkSj ; esa oqQN vkarfjd fodYi iznku fd, x, gSaA vkidks
,sls iz'uksa esa ls osQoy ,d fodYi dk mÙkj nsuk gSA
10. oSQYD;qysVj osQ iz;ksx dh vuqefr ugha gSA
General Instructions:
1. All questions are compulsory.
2. There are 35 questions in all.
3. This question paper has five sections. Section-A, Section-B, Section-C, Section-D and
Section-E. All these Sections are compulsory.
4. Section-A contain 18 objective type questions of 1 mark each.
5. Section-B contain 7 very short answer type questions of 2 marks each.
6. Section-C contain 5 short answer type questions of 3 marks each.
7. Section-D contain 3 long answer type questions of 5 marks each.
8. Section-E contain 2 case study based questions of 4 marks each.
9. There is no overall choice. However, some internal choice has been provided in Section B,
C, D and E. You have to attempt only one of the choice in such questions.
10. Use of calculator is not permitted.

section-A

1. 436.32, 227.2 ,oa 0.301 la[;kvksa dk ;ksx mi;qDr lkFkZd vadksa esa gSµ1
(d) 663.821 ([k) 664 (x) 663.8 (?k) 663.82

The sum of numbers 436.32, 227.2 and 0.301 in appropriate significant figures is:

(a) 663.821 (b) 664 (c) 663.8 (d) 663.82

Page-1

2. ;fn laosx (P)] {ks=kiQy (A) ,oa le; (T) dks ewy jkf'k;k¡ eku ysa rks ÅtkZ dk foeh; lw=k gksxkµ1
(d) [P1A–1T1] ([k) [P2A1T1] (x) [P1A–1/2T1] (?k) [P1A1/2T –1]
If momentum (P), area (A) and time (T) are taken to be fundamental quantities, then energy

has dimensional formula:
(a) [P1A–1T1] (b) [P2A1T1] (c) [P1A–1/2T1] (d) [P1A1/2T –1]
3. d.kksa osQ chp la?kV~V esa laosx laj{k.k dk vocks/u fdl vk/kj ij fd;k tk ldrk gS\1
(d) ÅtkZ laj{k.k ([k) osQoy U;wVu dk izFke fu;e
(x) osQoy U;wVu dk f}rh; fu;e (?k) U;wVu dk f}rh; ,oa r`rh; fu;e
Conservation of momentum in a collision between particles can be understood from:

(a) Conservation of energy (b) Newton’s first law only
(c) Newton’s second law only (d) Both Newton’s second and third law
4. tehu ij mdMw cSBk gqvk O;fDr mBdj lh/k [kM+k gksrk gSA bl izfØ;k esa O;fDr ij yxus okyk i`Foh dk
izfrfØ;k cyµ1
(d) vifjo£rr jgrk gS vkSj ifjek.k esa mg osQ cjkcj gksrk gSA
([k) vifjo£rr jgrk gS vkSj ifjek.k esa mg ls vf/d gksrk gSA
(x) izkjaHk ifjorhZ ijarq ifjek.k esa lnSo mg ls vf/dA
(?k) izkjaHk esa mg ls vf/d gksrk gS ijarq ckn esa mg osQ cjkcj gks tkrk gSA
A man squatting on the ground gets straight up and stand. The force of reaction of ground

on the man during the process is:
(a) constant and equal to mg in magnitude.
(b) constant and greater than mg in magnitude.
(c) variable but always greater than mg.
(d) at first greater than mg, and later becomes equal to mg.
5. fp=k esa n'kkZ, x, fudk; esa vafdr dkSu&lk fcanq blosQ æO;eku [kks[kyk xksyk
ok;q
osaQæ dh laHkkfor fLFkfr gS\1
(d) A A
R/2

([k) B B
C
(x) C
R/2
(?k) D D
jsr
Which of the following points is the likely position of the centre of mass of the system

shown in figure. Hollow
Air
(a) A sphere

(b) B R/2 A

(c) C B
C
(d) D R/2
Sand
D

Page-2

6. fdlh vk;rkdkj izsQe dks nks leku yackbZ dh Mksfj;ksa }kjk nks voyacksa ls lefer :i ls fuyafcr fd;k tkuk
gSA bls uhps fn, x, fp=k esa rhu <+axksa ls fd;k tk ldrk gSA Mksjh esa rukoµ1

(d) ([k) (x)
(d) lHkh izdj.kksa esa leku gksxk ([k) (d) esa lcls de gksxk
(x) ([k) esa lcls de gksxk (?k) (x) esa lcls de gksxk
A rectangular frame is to be suspended symmetrically by two strings of equal length on

two supports as shown in Fig. It can be done in one of the following three ways:

(a) (b) (c)
The tension in the strings will be:
(a) the same in all cases (b) least in (a)
(c) least in (b) (d) least in (c)
7. fdlh ekè;e esa vuqnSè;Z rjaxksa osQ izxeu ls tks jkf'k laapfjr gksrh gS og gSµ1
(d) æO; ([k) ÅtkZ (x) ÅtkZ ,oa æO; (?k) ÅtkZ] æO; ,oa laosx
With propagation of longitudinal waves through a medium, the quantity transmitted is:

(a) matter (b) energy
(c) energy and matter (d) energy, matter and momentum
8. fdlh vkn'kZ xSl osQ fy, fn, x, æO;eku osQ fy,] nkc osQ nks fHkUu ekuksa V
(l)
osQ fy,] vk;ru ,oa rki osQ chp xzkiQ n'kkZ;k x;k gSA P1 rFkk P2 osQ chp P2

laca/ osQ ckjs esa D;k fu"d"kZ fudkyk tk ldrk gS\1 P1
(d) P1 > P2
([k) P1 = P2
(x) P1 < P2
T (k)
(?k) vkadM+s i;kZIr ugha gSa
Volume versus temperature graphs for a given mass of an

ideal gas are shown in Fig. at two different values of constant P2
pressure. What can be inferred about relation between P1 and
V (l)

P1
P2?
(a) P1 > P2
(b) P1 = P2
(c) P1 < P2
(d) data is insufficient T (k)

Page-3

9. dksbZ vkn'kZ xSl fp=k esa P-V vkjs[k osQ vuqlkj pØh; izfØ;k ABCDA djrh gSA
xSl }kjk fd, x, dk;Z dh ek=kk gSµ1
P

D C
2PO

PO
A B

VO 3VO V

(d) 6P0V0 ([k) –2P0V0 (x) +2P0V0 (?k) +4P0V0
An ideal gas undergoes cyclic process ABCDA as shown in P-V diagram.

P

D C
2PO

PO
A B

VO 3VO V

The amount of work done by the gas is:
(a) 6P0V0 (b) –2P0V0 (c) +2P0V0 (d) +4P0V0
10. fdlh lfn'k dk ifjek.k lnSo ,d ----------------------- gksrk gSA1
The magnitude of a vector is always a ..................... .
11. ,d fu;fer o`rkdkj oy; osQ O;kl osQ v{k ij tM+Ro vk/w.kZ -------------------- gksrk gSA1
Moment of inertia of a uniform circular ring about a diameter of a ring is .................... .
12. O;kid :i ls æoksa dk i`"B ruko rki c<+us ij ----------------- gSA1
Surface tension of liquid generally .................... with temperature.

13. fdlh d.k dk vkSlr osx blosQ rkR{kf.kd osx osQ cjkcj gSA bldh xfr fdl izÑfr dh gS\1
The average velocity of a particle is equal to its instantaneous velocity. What is the nature
of its motion.
14. æo dk vi:i.kkad fdruk gksrk gS\1
Give value for the modulus of rigidity of liquids.
15. /w.kZu xfr esa fd;s x;s dk;Z osQ fy, lw=k fyf[k,A1
Write down the formula for work done in rotatory motion.
fuEufyf[kr izR;sd iz'u (izú laú 16, 17 rFkk 18) esa nks dFku fn, x, gSµ vfHkdFku (A) rFkk dkj.k (R)]
uhps fn, x, owQV % (A), (B), (C) vkSj (D) ls bu iz'uksa dk lgh mÙkj dk p;u djsa%

Page-4

(A) A vkSj R nksuksa lR; gS vkSj R, A dh lgh O;k[;k gSA
(B) A vkSj R nksuksa lR; gSa] ysfdu R, A dh lgh O;k[;k ugha gSA
(C) A lR; gS] ysfdu R vlR; gSA
(D) A vkSj R nksuksa vlR; gSA
Two statements are given below in each question (Q. No. 16, 17 and 18); one labelled Assertion
(A) and other labelled Reason (R). Select the correct answer to these questions from the codes:
(A), (B), (C) and (D) are as given below:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A and R both are false.
16. vfHkdFku (A) % leku Roj.k okys fiaM (oLrq) ges'kk ,d lh/h js[kk esa xfr djrs gSaA1
dkj.k (R) % lh/h js[kk esa xfr djuk (fiaM) oLrq dh LokHkkfod izo`fr gSA
Assertion (A) : When a body is subjected to a uniform acceleration, it always moves in a
straight line.
Reason (R) : Straight line motion is the natural tendency of the body.
17. vfHkdFku (A) % coaMj esa ,d /werh gqbZ gok dh xfr [krjukd :i ls vf/d gksrh gSA1
dkj.k (R) % ;fn fdlh fiaM ij dksbZ ckgjh cy vk?kw.kZ dk;Z ugha djrk gS] rks mldk dks.kh; laosx
fLFkj jgrk gSA
Assertion (A) : The speed of a whirl wind in a tornado is alarmingly high.
Reason (R) : If no external torque acts on a body, its angular velocity remains constant.
18. vfHkdFku (A) % rkieku esa o`f¼ osQ lkFk ,d Bksl vkSj æo dk laioZQ dks.k c<+rk gSA1
dkj.k (R) % rkieku esa o`f¼ osQ lkFk] æo dk i`"B ruko vf/d gks tkrk gSA
Assertion (A) : The angle of contact of a liquid with a solid decreases with increase in
temperature.
Reason (R) : With increase in temperature, the surface tension of the liquid increases.
section-B

19. foek fof/ dk mi;ksx djosQ 10 twy dks vxZ esa ifjo£rr djsaA 2
Convert 10 Joule into erg, using dimensional method.
vFkok (Or)
A
lw=k  P + 2  (V − B) = RT, esa P nkc gS vkSj V vk;ru gSA
 V 
A vkSj B dh foek Kkr dhft,A
 A 
In the formula  P + 2  (V − B) = RT.
 V 
Where P is pressure and V is the volume. Calculate the dimensions of A and B.
20. canwd dk izfr{ksi osx D;k gksrk gS\ canwd osQ izfr{ksi osx osQ fy, O;atd izkIr dhft,A 2
What is recoil velocity of gun? Obtain an expression for recoil velocity.


Page-5

vFkok (Or)
3 kg lagfr osQ fdlh fi.M ij vkjskfir dksbZ cy 25 s esa mldh pky dks 2 ms–1 ls 3.5 ms–1 dj nsrh
gSA fi.M dh xfr dh fn'kk vifjo£rr jgrh gSA cy dk ifjek.k o fn'kk D;k gS\
A constant force acting on a body of mass 3 kg changes its speed from 2 ms–1 to 3.5 ms–1 in

25 sec. The direction of the motion of the body remains unchanged. What is the magnitude
and direction of the force?
21. /ukRed o ½.kkRed dk;Z D;k gksrk gS\ mnkgj.k lfgr crkb,A 2
What do you mean by positive and negative work done? Explain with example.

vFkok (Or)
vleku æO;eku osQ nks fiaM ,d gh fn'kk esa leku xfrt ÅtkZ ls xfreku gSaA nksuks fiaMksa ij cjkcj ifjek.k
dk cy yxkdj mUgsa fojke voLFkk esa yk;k x;k gSA fojkekoLFkk esa vkus rd muosQ }kjk pyh xbZ nwfj;ksa dh
rqyuk dhft,A
Two bodies of unequal mass are moving in the same direction with equal kinetic energy.
The two bodies are brought to rest by applying retarding force of same magnitude. How
would the distance moved by them before coming to rest? Compare.
22. /w.khZ xfr esa æO;eku osQ lerqY; jkf'k D;k gS\ ml jkf'k dks ifjHkkf"kr dhft,\ 2
What is the analogue of mass in rotational motion? Define that physical quantity.

23. osQIyj }kjk fn, x, (i) d{kkvksa dk fu;e o (ii) {ks=kiQyksa dk fu;e dh ifjHkk"kk fyf[k,A 2
State (i) law of orbits and (ii) law of areas given by Kepler for planetry motion.

24. lerkih; izØe vkSj :}h"e izØe esa dksbZ nks varj crkb,\ 2
Given any two differences between isothermal process and an adiabatic process.

25. fdlh d.k dh ljy vkorZ xfr esa Roj.k osQ fy, lw=k Kkr dhft,A Roj.k o vkorZdky (T) dk vkjs[k
cukb,A 2
Find expression for acceleration of a particle executing simple harmonic motion. Draw a

graph between acceleration and time period (T).
section-C

26. m"ekxfrdh osQ izFke fu;e dh lgk;rk ls es;j osQ lw=k CP – CV = R dk fuxeu dhft,A 3
Establish Mayer’s Formula CP – CV = R by using first law of thermodynamics.

vFkok (Or)
(i) D;k fdlh xSl dks fcuk m"ek fn, mlosQ rki esa o`f¼ dh tk ldrh gS\ Li"V dhft,A
(ii) vkarfjd ÅtkZ o rki dh vo/kj.kk m"ekxfrdh osQ dkSu ls fu;eksa }kjk fodflr dh xbZ gS\
(i) Is it possible to increase the temperature of a gas without adding heat to it? Explain.
(ii) Which law of thermodynamics has been given the concept of internal energy and
temperature.
27. ,d ';ku æo esa fxjrh xksyh osQ fy, lhekar osx dk O;atd Kkr djsaA 3
Obtain an expression for terminal velocity of a ball falling in a viscous liquid.
vFkok (Or)
izfrcy&foÑfr oØ dh lgk;rk ls rU;] Hkaxqj vkSj izR;LFkkyd esa vrj Li"V dhft,A
On the basis of stress-strain curve, differentiate between ductile, brittle and elastomers.


Page-6

28. ,dijek.kqd xSal osQ fy, fof'k"V&m"ek /kfjrk dk O;atd Kkr dhft,A 3
Calculate specific heat capacity of a monoatomic gas.

29. fdlh fLizax dh izR;kLFk fLFkfrt ÅtkZ osQ fy, O;atd Kkr dhft,A 3
Derive expression for elastic potential energy of a spring.
30. lfn'k ;ksx dk f=kHkqt fu;e crkb, vkSj ifj.kke osQ fy, lw=k LFkkfir dhft,A 3
State triangle law of vector addition. Find out the expression for resultant vector.

section-D

31. ljy yksyd osQ vkoZrdky osQ fy, O;atd izkIr dhft,A 5

Obtain an expression for the time-period of a simple pendulum.

vFkok (Or)
(i) èofu osQ osx osQ fy, U;wVu osQ lw=k dk O;atd izkIr djsaA
(ii) 300 m Å¡ph ehukj osQ 'kh"kZ ls fxjk;k x;k iRFkj ehukj osQ vk/kj ij cus rkykc osQ ikuh ls Vdjkrk
gSA ;fn ok;q esa èofu dh pky 340 ms–1 gS rks iRFkj osQ Vdjkus dh èofu ehukj osQ 'kh"kZ ij iRFkj
fxjkus osQ fdruh nsj ckn lqukbZ nsxh\ (g = 9.8 ms–2)
(i) Obtain an expression for Newton’s formula for velocity of sound in gases.
(ii) A stone dropped from the top of a tower of height 300 m splashes into the water of
a pond near the base of the tower. When is the splash heard at the top given that the
speed of sound in air is 340 ms–1? (g = 9.8 ms–2)
32. osQf'kdk uyh esa æo osQ osQf'kdh; mUu;u (h) osQ fy, O;atd izkIr dhft,A 5
Derive expression for height of the liquid rise (h) in a capillary tube.
vFkok (Or)
U;wVu osQ 'khryu osQ fu;e dks fy[ksa vkSj fl¼ djsaA
State and prove Newton’s law of cooling.
33. ,d iz{ksI; dks {kSfrt osQ lkFk q dks.k ij osx u osQ lkFk iz{ksfir fd;k tkrk gSA
(d) n'kkZb, fd bldk iz{ksI; iFk ,d i[ky; gSA

([k) fuEufyf[kr osQ fy, O;atd izkIr dhft,+%
(i) mM~M;u dk le;
(ii) izkIr vf/dre Å¡pkbZ 5
A projectile is fired at an angle q with the horizontal with velocity ‘u’.
(a) Show that its trajectory is parabolic.
(b) Obtain an expression for;
(i) Time of flight
(ii) Maximum height attained.5
vFkok (Or)
(i) fdlh jktekxZ ij iqfyl dh dksbZ xkM+h 30 km/h dh pky ls py jgh gS vkSj ;g mlh fn'kk esa
192 km/h dh pky ls tk jgh fdlh pksj dh dkj ij xksyh pykrh gSA ;fn xksyh dh uky eq[kh pky
150 ms–1 gS rks pksj dh dkj dks xksyh fdl pky osQ lkFk vk/kr djsxhA
(ii) dyu&fof/ dk mi;ksx djosQ ,d leku Roj.k osQ fy, v = u + at lehdj.k izkIr dhft,A
(i) A police van moving on a highway with a speed of 30 km/h fires a bullet at a thief’s

Page-7

car speeding away in the same direction with a speed of 192 km/h. If the muzzle
speed of the bullet is 150 ms–1, with what speed does the bullet hit the thief’s car?
(ii) Obtain the equation of motion v = u + at for constant acceleration using method of
calculus.
section-E Case Study

34. ge izk;% ?k"kZ.k dks ,d vokaNuh; cy ekurs gSaA cgqr lh fLFkfr;ksa es] tSls fdlh e'khu] ftlesa fofHkUu
dy iqtsZ xfr djrs gksa] esa ?k"kZ.k dh ½.kkRed Hkwfedk gksrh gSA ;g vkis{k xfr;ksa dk fojks/ drk gS ftlosQ
iQyLo:i m"ek vkfn osQ :i esa ÅtkZ&{k; gksrh gSA rFkkfi] cgqr&lh O;kogkfjd fLFkfr;ksa esa] ?k"kZ.k vR;Ur
vko';d gksrk gSA xfrt ?k"kZ.k esa ÅtkZ&{k; gksrh gS] fiQj Hkh vkisf{kd xfr dks 'kh?kz lekIr djus esa bldh
egRoiw.kZ Hkwfedk gSA e'khuksa rFkk ;a=kksa esa czsd dh Hkkafr bldk mi;ksx fd;k tkrk gSA
(i) ?k"kZ.k cy gS% 1
(a) ,d laj{kh cy (b) ,d Nn~e cy (c) ,d vlaj{kh cy (d) vfHkosQUæ cy
(ii) dkSu&lk ?k"kZ.k vf/dre gS\ 1
(a) LFkSfrd ?k"kZ.k (b) lhekUr ?k"kZ.k (c) xfrt ?k"kZ.k (d) dksbZ ugha
(iii) cky fc;fjax dk iz;ksx djosQ lihZ ?k"kZ.k dks ---------------- ?k"kZ.k esa ifjo£rr fd;k tk ldrk gSA 1
(iv) ?k"kZ.k dks.k vkSj ?k"kZ.k xq.kkad esa D;k laca/ gS\ 1
OR
?k"kZ.k dks de oSQls fd;k tk ldrk gSA
We often regard friction as something undesirable. In many situations, like in a machine

with different moving parts, friction does have a negative role. It opposes relative motion
and thereby dissipates power in the form of heat. In many practical situations, however
friction is critically needed. Kinetic friction that dissipates power is nevertheless important
for quickly stopping relative motion. It is made use of the brakes in machines and
automobiles.
(i) Friction forces is: 1
(a) a conservative force (b) Pseudo force
(c) a non-conservative force (d) centripetal force
(ii) Which friction is maximum? 1
(a) Static friction (b) Limiting friction (c) Kinetic friction (d) None
(iii) By using ball bearings, sliding friction can be converted into .............. friction.
(iv) What is relation between angle of friction and coefficient of friction? 1
OR
How can we reduced friction.
1
35. ge tkurs gSa fd i`Foh oLrqvksa dks viuh vksj vkd£"kr djrh gSA i`Foh osQ bl vkd"kZ.k cy dks xq#Roh; cy
dgrs gSaA vr% tc oLrq,¡ i`Foh dh vksj osQoy blh cy osQ dkj.k fxjrh gS] ge dgrs gSa fd oLrq,¡ eqDr
iru esa gSaA D;k fxjrh gqbZ oLrqvksa osQ osx esa dksbZ ifjorZu gksrk gS\ fxjrs le; oLrqvksa dh xfr dh fn'kk esa
dkbZ ifjorZu ugha gksrkA ysfdu i`Foh osQ vkd"kZ.k osQ dkj.k osx osQ ifjek.k esa ifjorZu gksrk gSA osx esa dksbZ
Hkh ifjorZu Roj.k dks mRiUu djrk gSA tc Hkh dksbZ oLrq i`Foh dh vksj fxjrh gS] Roj.k dk;Z djrk gSA ;g
Roj.k i`Foh osQ xq#Roh; cy osQ dkj.k gSA xq#Rokd"kZ.k osQ dkj.k Roj.k osQ ckjs esa mYys[kuh; ckr ;g gS fd
;g viuh lrg ij vf/dre gksrk gS] pkgs vki Åij tk,a ;k uhps] ;g ?kVrk tkrk gSA

Page-8

(i) eqDr :i ls fxjus osQ nkSjku fdlh fiaM dk xq#Rokd"kZ.k osQ dkj.k Roj.k fdl ij fuHkZj ugha djrk gS% 1
(a) i`Foh dk æO;eku (b) fxjus okys fiaM dk æO;eku
(c) lkoZf=kd xq#Rokd"kZ.k fLFkjkad (d) i`Foh dh f=kT;k
(ii) tc fdlh fiaM dks Åij isaQdk tkrk gS] rks xq#Rokd"kZ.k cy fdl fn'kk esa yxrk gS\ 1
(a) Åij dh vksj (b) uhps dh vksj (c) 'kwU; (d) {kSfrt fn'kk esa
(iii) ,d d.k dks i`Foh dh lrg ls R Å¡pkbZ rd ys tk;k tkrk gS] tgk¡ R i`Foh dh f=kT;k gSA xq#Rokd"kZ.k
osQ dkj.k Roj.k gS% 1
(a) 2.45 ms–2 (b) 4.9 ms–2 (c) 9.8 ms–2 (d) 19.6 ms–2
(iv) ml oLrq dk æO;eku D;k gS\ ftldk Hkkj i`Foh ij 49 N gS\ 1
OR
xgjkbZ c<+us osQ lkFk xq#Rokd"kZ.k osQ dkj.k Roj.k -------------------- gSA
We have learnt that the earth attracts objects towards it. This is due to the gravitational

force. Whenever objects fall towards the earth under this force alone, we say that the objects
are in free fall. Is there any change in the velocity of falling objects? While falling, there is
no change in the direction of motion of the objects. But due to earth’s attraction, there will
be a change in the magnitude of velocity. Any change in the velocity involves acceleration.
Whenever an object falls towards the earth, an acceleration is involved. This acceleration is
due to earth’s gravitational force. The remarkable thing about acceleration due to gravity
is that it is maximum on its surface, decreasing whether you go up or down.
(i) Acceleration due to gravity of a body during free-fall does not depend upon the: 1
(a) mass of Earth (b) mass of falling body
(c) universal gravitational constant (d) radius of Earth
(ii) When a body is thrown up, the force of gravity is: 1
(a) in the upward direction (b) in the downward direction
(c) zero (d) in the horizontal direction
(iii) A particle is taken to a height R above the Earth’s surface, where R is the radius of
Earth. The acceleration due to gravity there is: 1
(a) 2.45 ms–2 (b) 4.9 ms–2 (c) 9.8 ms–2 (d) 19.6 ms–2
(iv) What is the mass of an object whose weight is 49 N on the Earth? 1
OR
Acceleration due to gravity ................... with increasing depth.
 qqq

Page-9