JEE (Main + Advanced) 2026
JEE (Main + Advanced) 2026
NURTURE COURSE
NURTURE COURSE
RACE # 21 PHYSICS
1. In the given figure choose the correct statement regarding the acceleration of masses just after the string
between 'A' and ground is cut.
(A) All the three blocks move with same acceleration æç ö÷ .
g
è3ø
g
(B) Acceleration of aC = aB = 0 and aA = . (B) M
2
M (A)
g
(C) aA = aB = ; aC = 0. (C) M
2
g
(D) aA = aB = ; a = 0.
3 C
iznf'kZr fp= esa /kjkry rFkk 'A' ds e/; yxh jLlh dks dkVus ds rqjUr i'pkr~ æO;ekuksa ds Roj.k ds ckjs esa lgh dFku
pqfu,%&
ægö
(A) lHkh rhuksa CykWd leku Roj.k ç ÷ ds lkFk xfr djsaxAs
è3ø
g
(B) Roj.k aC = aB = 0 rFkk aA = gS (B) M
2
M (A)
g
(C) aA = aB = ; aC = 0 (C) M
2
g
(D) aA = aB = ;a =0
3 C
Ans. (C)
2. A rope whose mass per unit length varies according to the relation l = 4x [kg/m] is being pulled by a
constant horizontal force of 10N over a smooth floor (lenght of rope, L = 1m) as shown in the figure.
Find
(a) acceleration of the rope
L
(b) tension at point P if x = .
4
,d jLlh dk çfr bdkbZ yEckbZ dk æO;eku laca/k l = 4x [kg/m] ds vqulkj ifjofrZr gksrk gS rFkk bls ,d fpdus Q'kZ
ij 10N ds fu;r {kSfrt cy }kjk [khapk tkrk gAS (jLlh dh yEckbZ L = 1m)
(a) jLlh dk Roj.k
L
(b) fcUnq P ij ruko Kkr dhft;s ;fn x = gksA
4
Ans. (a) 5 m/s2, (b) 5/8 N
3. A block of mass 2kg is at rest on a rough surface, of coefficient of friction 0.5. A force F is applied on
the block, variation of force is represented in the figure. Speed of the block at t = 4sec is :
æO;eku 2kg okyk ,d CykWd ?k"kZ.k xq.kkad 0.5 okyh [kqjnjh lrg ij fojkekoLFkk esa j[kk gAS CykWd ij ,d cy F yxk;k
tkrk gS ftlesa ifjorZu fp= esa n'kkZ;k x;k gAS t = 4 lsd.M ij CykWd dh pky gksxh %&
PHY. / R # 21 E-1/5
JEE (Main + Advanced) 2026
NURTURE COURSE
5N
M = 0.5 F
1sec t
(A) 20 m/s (B) 10 m/s (C) 5 m/s (D) 15 m/s
Ans. (C)
Sol. F = 10N at t = 2 sec
a Fnet
5N
t=2sec 4sec t 2sec t
1
v–u= ×2×5
2
v = 5 m/s
4. The system shown is just on the verge of slipping . The co-efficient of static friction between the block
and the table top is:
fp= esa iznf'kZr fudk; fQlyus ds Bhd djhc gAS CykWd rFkk Vscy ds 'kh"kZ ds e/; LFkSfrd ?k"kZ.k xq.kkad gksxk%&
(A) 0.5 (B) 0.95 (C) 0.15 (D*) 0.35
Ans. (D)
5. A block of mass M placed on rough surface of coefficient of friction equal to 3. If F is the (4/5) of the
minimum force required to just move, find out the force exerted by ground on the block.
M nzO;eku dk ,d CykWd] ?k"kZ.k xq.kkad 3 okyh ,d [kqjnjh lrg ij fLFkr gAS ;fn F dk eku] CykWd dks xfr djkus ds
fy;s vko';d U;wure cy dk (4/5) gks rks /kjkry }kjk CykWd ij vkjksfir cy Kkr dhft;sA
M F
m=3
(A) 2.6 Mg (B) Mg (C) 4 Mg (D) 3.4 Mg
Ans. (A)
Sol. The force exerted by ground on the block
= N2 + f 2
2
é
æ 4ö ù 13
= ( Mg) + êçè 5 ÷ø ( 3Mg ) ú
2
= Mg = 2.6Mg
ë û 5
E-2/5 PHY. / R # 21
JEE (Main + Advanced) 2026
NURTURE COURSE
6. The coefficient of friction between the block and the horizontal surface is µ as shown in figure. The
block moves towards right under action of horizontal force F(figure -a). Sometime later another force P
is applied to the block making an angle q (such that tanq = µ) with vertical as shown in (figure - b). After
application of force P, the acceleration of block shall
P
m q
m F m F
Fig. (a) Fig. (b)
(A) increase (B) decrease
(C) remains same (D) information insufficient for drawing inference.
fp= esa iznf'kZr CykWd rFkk {kSfrt lrg ds e/; ?k"kZ.k xq.kkad µ gSA ;g CykWd fp= (a) ds vuqlkj ,d {kfS rt cy F ds
izHkko esa nka;h vksj xfr djrk gAS dqN le; i'pkr~ fp= (b) ds vuqlkj bl CykWd ij ,d vU; cy P Å/okZ/kj ls dks.k
q (tanq = µ) ij yxk;k tkrk gAS cy P yxkus ds i'pkr~ CykWd dk Roj.k %&
P
m q
m F m F
Fig. (a) Fig. (b)
(A) c<+ tkrk gAS (B) ?kV tkrk gAS (C) ogh jgrk gSA (D) vkdM+s vi;kZIr gAS
Ans. (C)
Sol. Pcosq Psinq
Q m P cos q = P sin q
So acceleration of block remains same.
7. For shown situation the acceleration of block is represented by : (Here t is time in second) (g = 10ms–2)
n'kkZ;h xbZ fLFkfr ds fy;s CykWd ds Roj.k dks iznf'kZr fd;k tk ldrk gS (;gk¡ t lsd.M esa le; gS ) (g = 10ms–2)
m s=0.4 10kg F=2t
m k=0.3
a a
tan–1(0.2)
(A) –1 (B) 1
tan (0.2)
t(s) t(s)
0 10 20 0 10 20
tan–1(0.1)
(C) 1 (D) None of these / buesa ls dksbZ ugha
t(s)
0 10 20
Ans. (B)
40 - 30
Sol. Block moves of 2t ³ ( 0.4)(10)(10) Þ t ³ 20s ; Just after start its acceleration a = = 1ms -2
10
2t - 30 t æ 1ö
After that a = = - 3 Þ Slope of a–t curve tan -1 ç ÷ = tan -1 ( 0.2)
10 5 è 5ø
PHY. / R # 21 E-3/5
JEE (Main + Advanced) 2026
NURTURE COURSE
8. A block is moving along y–axis with velocity vr A = 4ˆj on a plank relative to ground and the plank is
r
moving along x–axis with velocity v P = 3iˆ at any instant of time. The unit vector in direction of friction
force acting on block at this moment of time is
r r
,d CykWd v = 4ˆj osx ls /kjkry ds lkis{k y–v{k ds vuqfn'k r[rs ij xfr dj jgk gAS r[rk fdlh {k.k v = 3iˆ
A P
osx ls x– v{k ds vuqfn'k xfr dj jgk gSA bl {k.k ij CykWd ij dk;Zjr ?k"kZ.k cy dh fn'kk esa ,dkad lfn'k gksxk%&
y
x
A
3iˆ + 4ˆj 3iˆ - 4ˆj -3iˆ + 4 ˆj -3iˆ - 4 ˆj
(A) (B) (C) (D)
5 5 5 5
Ans. (B)
ˆ é 4ˆj - 3iˆ ù 3iˆ - 4jˆ
Sol. f = - vˆ rel = - vˆ AP = - ê ú=
ë 5 û 5
9. For the system of blocks and pulley shown in the figure, acceleration of block is "a" and friction force
acting on 10 kg block is F then :-
fp= esa iznf'kZr CykWd f?kjuh fudk; esa CykWd dk Roj.k a rFkk 10 kg CykWd ij dk;Zjr ?k"kZ.k cy F gks rks %&
10
µ = 0.6
20
(A) F = 60 N, a = 0 m/s2 (B) F = 50 N, a = m/s2
7
20
(C) F = 40 N, a = 0 m/s2 (D) F = 40 N, a = m/s2
7
Ans. (C)
Sol. In the given situation the blocks will not be able to move, i.e. T = 40 N
Thus, friction = 40 N
&a=0
10. A block of mass 2 kg is pressed against the vertical wall of coefficient of friction m = 0.2. The value of
friction force acting on the wall is :-
(A) 40 3 N in upward direction (B) 40 3 N in downward direction
(C) 20N in upward direction (D) No friction force acts on the block
E-4/5 PHY. / R # 21
JEE (Main + Advanced) 2026
NURTURE COURSE
m=0.2
2 kg
60°
40N
2 kg nzO;eku ds ,d CykWd dks m = 0.2 ?k"kZ.k xq.kkad okyh Å/okZ/kj nhokj ij nck;k tkrk gSA nhokj ij dk;Zjr~ ?k"kZ.k
cy dk eku g&S
(A) Å
¥ ij dh fn'kk esa 40 3 N (B) uhps dh fn'kk esa 40 3 N
(C) Åij dh fn'kk esa 20N (D) CykWd ij dksbZ ?k"kZ.k cy dk;Z ugha djrk gAS
Ans. (D)
20 3 2kg
Sol. As net force along the surface is zero there is no need of friction \fr = 0
20
20
11. Figure shows a block kept on a rough inclined plane. The maximum external force down the incline for
which the block remains at rest is 2N while the maximum external force up the incline for which the
block is at rest is 10 N. The coefficient of static friction m is
fp=kuqlkj ,d CykWd ,d [kqjnjs ur&ry ij j[kk gqvk gAS CykWd dks fojkekoLFkk esa cuk;s j[kus ds fy;s ur&ry ij uhps
dh vksj yxk;k tkus okyk vf/kdre ckg~; cy 2 N gS tcfd bls fojkekoLFkk esa j[kus ds fy;s ur&ry ij Åij dh
fn'kk esa yxk;k tkus okyk vf/kdre ckg~; cy 10 N gSA LFkSfrd ?k"kZ.k xq.kkad m dk eku gksxk
30°
3 1 1
(A) (B) (C) 3 (D)
2 6 3
Ans. (A)
Sol. When block is about to slide down 2 + mg sin q = m mg cos q ....(1)
3
when block is about to slide up mg sin q + m mg cos q = 10 ....(2) from (1) and (2) m =
2
N_Race # 21 ANSWER KEY
1. Ans. (C) 2. Ans. (a) 5 m/s2, (b) 5/8 N 3. Ans. (C)
4. Ans. (D) 5. Ans. (A) 6. Ans. (C) 7. Ans. (B)
8. Ans. (B) 9. Ans. (C) 10. Ans. (D) 11. Ans. (A)
PHY. / R # 21 E-5/5
