jee main question papers & answer keys

395 Pages·2016·7.32 MB·English

JEE MAIN QUESTION
PAPERS & ANSWER
KEYS (2014-2017)
Introduction
Careers360 brings you a compilation of JEE Main Online Question Papers and Answer Keys from
the year 2014 to 2017. This e-book helps you master questions which appeared in the national
level engi-neering entrance examination in the last 4 years.
JEE Main 2017 Official Question Paper 1 Physics, April 9
(English+Hindi)
PHYSICS ÷ÊÒÁÃ∑§ ÁflôÊÊŸ
1. A physical quantity P is described by the 1. ∞∑§ ÷ÊÒÁÃ∑§ ⁄UÊÁ‡Ê P ÁŸêŸ ‚¥’¥œ mÊ⁄UÊ ¬Á⁄U÷ÊÁ·Ã ∑§Ë
relation
¡ÊÃË „Ò–
1
P = a 2 b2 c3 d−4
1
P = a 2 b2 c3 d−4
If the relative errors in the measurement
ÿÁŒ •ÊÒ⁄U ∑§ ◊ʬŸ ◊¥ ‚ʬˇÊ òÊÈÁ≈U ∑˝§◊‡Ê—
of a, b, c and d respectively, are 2%, 1%, a, b, c d
3% and 5%, then the relative error in P fl „Ê ÃÊ ◊¥ ‚ʬˇÊ òÊÈÁ≈U „ÊªË —
2%, 1%, 3% 5% P
will be :
(1) 8%
(2) 12%
(1) 8%
(3) 32%
(2) 12%
(4) 25%
(3) 32%
(4) 25%
2. A car is standing 200 m behind a bus,
which is also at rest. The two start moving
at the same instant but with different 2. ∞∑§ ∑§Ê⁄U ∞∑§ ∆U„⁄UË „ÈÿË ’‚ ‚ 200 m ¬Ë¿U π«∏Ë „Ò–
forward accelerations. The bus has ŒÊŸÊ¥ ∞∑§ „Ë ˇÊáÊ •‹ª-•‹ª •ª˝ÁŒÁ‡Ê∑§ àfl⁄UáÊ ‚
2
acceleration 2 m/s and the car has
ø‹ŸÊ ‡ÊÈM§ ∑§⁄UÃ „Ò¥– ’‚ ∑§Ê àfl⁄UáÊ 2 ÃÕÊ
2 2 m/s
acceleration 4 m/s . The car will catch
∑§Ê⁄U ∑§Ê àfl⁄UáÊ 2 „Ò– Á∑§ÃŸ ‚◊ÿ ’ÊŒ ÿ„
up with the bus after a time of : 4 m/s
∑§Ê⁄U ’‚ Ã∑§ ¬„È°ø ¡ÊÿªË?
(1) 110 s
(2) 120 s
(1) 110 s
(3) 10 2 s
(2) 120 s
(4) 15 s
(3) 10 2 s
(4) 15 s
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(English+Hindi)
3. Two particles A and B of equal mass M 3. ‚◊ÊŸ Œ˝√ÿ◊ÊŸ M ∑§ ŒÊ ∑§áÊ A ÃÕÊ B ‚◊ÊŸ
v
are moving with the same speed as øÊ‹v‚ ÁøòÊÊŸÈ‚Ê⁄U ø‹ ⁄U„ „Ò¥– fl„ ¬ÍáʸÃÿÊ •¬˝àÿÊSÕ

shown in the figure. They collide
‚¥ÉÊ^ ∑§⁄UÃ „Ò¥ ÃÕÊ ‚¥ÉÊ^ ∑§ ’ÊŒ ∞∑§ ∑§áÊ ∑§Ë Ã⁄U„
C
completely inelastically and move as a
single particle C. The angle θ that the path ø‹Ã „Ò¥– ∑§ÊáÊ θ, ¡Ê ∑§áÊ C ∑§Ê ¬Õ X-•ˇÊ ‚
of C makes with the X-axis is given by : ’ŸÊÃÊ „Ò, ∑§Ê ÁŸêŸ ‚ê’㜠‚ ÁŒÿÊ ¡ÊÿªÊ —
3+ 2 3+ 2
tanθ= tanθ=
(1) 1− 2 (1) 1− 2
3− 2 3− 2
tanθ= tanθ=
(2) 1− 2 (2) 1− 2
1− 2 1− 2
tanθ= tanθ=
(3) 2 (1+ 3) (3) 2 (1+ 3)
1− 3 1− 3
tanθ= tanθ=
(4) 1+ 2 (4) 1+ 2
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(English+Hindi)
4. The machine as shown has 2 rods of length 4. ÁøòÊ ◊¥ ÁŒπÊÿË ªÿË ∞∑§ ◊‡ÊËŸ ∑§Ë ŒÊ ¿U«∏Ê¥, Á¡Ÿ∑§Ë
1 m connected by a pivot at the top. The ‹ê’Ê߸ „Ò, ∑§ ™§¬⁄UË Á‚⁄UÊ¥ ∑§Ê ∞∑§ ‚ÊÕ œÈ⁄Uʪ˝SÃ
1 m
end of one rod is connected to the floor by
Á∑§ÿÊ ªÿÊ „Ò– ∞∑§ ¿U«∏ ∑§Ê •ÊÁπ⁄UË Á‚⁄UÊ ∞∑§ ÁSÕ⁄U
a stationary pivot and the end of the other
œÈ⁄UË mÊ⁄UÊ »§‡Ê¸ ‚ ¡Ê«∏Ê ªÿÊ „Ò ÃÕÊ ŒÍ‚⁄UË ¿U«∏ ∑§
rod has a roller that rolls along the floor in
a slot. As the roller goes back and forth, •ÊÁπ⁄UË Á‚⁄U ¬⁄U ∞∑§ ⁄UÊ‹⁄U ‹ªÊ „Ò ¡Ê Á∑§ »§‡Ê¸ ¬⁄U
a 2 kg weight moves up and down. If the
’Ÿ πÊ°ø ◊¥ ø‹ÃÊ „Ò– ¡’ fl„ ⁄UÊ‹⁄U •Êª ¬Ë¿U

roller is moving towards right at a constant
ø‹ÃÊ „Ò ÃÊ ∞∑§ ∑§Ê ÷Ê⁄U ™§¬⁄U ŸËø ø‹ÃÊ „Ò–
speed, the weight moves up with a : 2 kg
ÿÁŒ ⁄UÊ‹⁄U ŒÊÁ„ŸË ÁŒ‡ÊÊ ◊¥ ∞∑§ ‚◊ÊŸ øÊ‹ ‚ ø‹ÃÊ
„Ò ÃÊ fl„ ÷Ê⁄U ø‹ªÊ, ∞∑§ —
(1) constant speed
(2) decreasing speed
‚◊ÊŸ øÊ‹ ‚
(1)
(3) increasing speed
ÉÊ≈UÃË „È߸ øÊ‹ ‚
3 (2)
(4) speed which is th of that of the
4 ’…∏ÃË „È߸ øÊ‹ ‚
(3)
roller when the weight is 0.4 m
3
above the ground
øÊ‹ ¡Ê Á∑§ ⁄UÊ‹⁄U ∑§Ë øÊ‹ ∑§Ê „Ò ¡’ fl„
(4)
4
÷Ê⁄U »§‡Ê¸ ‚ ∑§Ë ™°§øÊ߸ ¬⁄U „Ò
0.4 m
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(English+Hindi)
5. A conical pendulum of length 1 m makes 5. ∞∑§ ‡ÊÊ¥∑§fl (conical) ŒÊ‹∑§, Á¡‚∑§Ë ‹ê’Ê߸ 1 m
an angle θ=458 w.r.t. Z-axis and moves „Ò •ÊÒ⁄U ¡Ê Z-•ˇÊ ‚ θ=458 ∑§ ∑§ÊáÊ ¬⁄U „Ò¥, XY
in a circle in the XY plane. The radius of
‚◊Ë ◊¥ ∞∑§ ªÊ‹Ê∑§Ê⁄U ¬Õ ◊¥ ø‹ÃÊ „Ò– ªÊ‹Ê∑§Ê⁄U
the circle is 0.4 m and its center is vertically
¬Õ ∑§Ë ÁòÊíÿÊ „Ò •ÊÒ⁄U ©‚∑§Ê ∑§ãŒ˝ Á’ãŒÈ
below O. The speed of the pendulum, in 0.4 m O
its circular path, will be : (Take g=10 ms−2) ∑§ ∆UË∑§ ŸËø „Ò– ©‚ ŒÊ‹∑§ ∑§Ë ªÁà ªÊ‹Ê∑§Ê⁄U ¬Õ
◊¥ „ÊªË — (g=10 ms−2)
(1) 0.4 m/s
(2) 4 m/s (1) 0.4 m/s
(3) 0.2 m/s (2) 4 m/s
(4) 2 m/s (3) 0.2 m/s
(4) 2 m/s
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(English+Hindi)
R 6. Œ˝√ÿ◊ÊŸ M ÃÕÊ ÁòÊíÿÊ R flÊ‹ ∞∑§‚◊ÊŸ Á«US∑§ ◊¥
6. A circular hole of radius is made in a
4 R
∞∑§ ÁòÊíÿÊ ∑§Ê ªÊ‹Ê∑§Ê⁄U ¿UŒ, ÁøòÊÊŸÈ‚Ê⁄U, Á∑§ÿÊ

thin uniform disc having mass M and 4
radius R, as shown in figure. The moment ªÿÊ „Ò– Á’¥ŒÈ ‚ ¡ÊŸ flÊ‹ ÃÕÊ Á«US∑§ ∑§ ‚◊Ë
O
of inertia of the remaining portion of the
∑§ ‹ê’flØ •ˇÊ ∑§ ‚ʬˇÊ, Á«US∑§ ∑§ ’ø „È∞ ÷ʪ
disc about an axis passing through the
∑§Ê, ¡«∏àfl •ÊÉÊÍáʸ „ÊªÊ —
point O and perpendicular to the plane of
the disc is :
2
219 MR
(1)
219 MR2 256
(1)
256
2
237 MR
(2)
237 MR2 512
(2)
512
2
19 MR
(3)
19 MR2 512
(3)
512
2
197 MR
(4)
197 MR2 256
(4)
256
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(English+Hindi)
7. The mass density of a spherical body is 7. ∞∑§ ªÊ‹Ê∑§Ê⁄U Á¬á«U ∑§Ê Œ˝√ÿ◊ÊŸ ÉÊŸàfl „Ò
k
given by ρ (r)= for r ≤ R and k
r ρ (r)= ¡’ r ≤ R ÃÕÊ
r
ρ (r)=0 for r > R,
where r is the distance from the centre. ρ (r)=0 ¡’ r > R, ¡„Ê° r ∑§ãŒ˝ ‚ ŒÍ⁄UË „Ò–
The correct graph that describes
qualitatively the acceleration, a, of a test ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ª˝Ê$»§ ∞∑§ ¬⁄UˡÊáÊ ∑§áÊ ∑§ àfl⁄UáÊ
particle as a function of r is : ∑§Ê ∑§ »§‹Ÿ ◊¥ ªÈáÊÊà◊∑§ M§¬ ‚ Œ‡ÊʸÃÊ „Ò?
a r
(1)
(1)
(2)
(2)
(3)
(3)
(4)
(4)
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(English+Hindi)
8. A steel rail of length 5 m and area of cross 8. 5 m ‹ê’Ê߸ ÃÕÊ 40 cm2 •ŸÈ¬˝SÕ ∑§Ê≈U ∑§ ˇÊòÊ»§‹
2
section 40 cm is prevented from expanding ∑§Ë ∞∑§ S≈UË‹ ∑§Ë ¬≈U⁄UË ∑§Ê ‹ê’Ê߸ ∑§ •ŸÈÁŒ‡Ê
along its length while the temperature rises
ÁflSÃÊ⁄UáÊ ⁄UÊ∑§Ê ¡ÊÃÊ „Ò ¡’Á∑§ ©‚∑§Ê Ãʬ◊ÊŸ 108C
by 108C. If coefficient of linear expansion
’…∏ÊÿÊ ¡ÊÃÊ „Ò– ÿÁŒ S≈UË‹ ∑§Ê ⁄UπËÿ ¬˝‚Ê⁄U ªÈáÊÊ¥∑§
and Young’s modulus of steel are
1.2×10−5 K−1 and 2×1011 Nm−2 ÃÕÊ ÿ¥ª ¬˝àÿÊSÕÃÊ ªÈáÊÊ¥∑§ ∑˝§◊‡Ê— 1.2×10−5 K−1
respectively, the force developed in the rail ÃÕÊ 2×1011 Nm−2 „Ò¥ ÃÊ ¬≈U⁄UË ◊¥ ©à¬ÛÊ ’‹ ∑§Ê
is approximately :
ÁŸ∑§≈UÃ◊ ◊ÊŸ „ÊªÊ —
7
(1) 2×10 N
5
(2) 1×10 N 7
(1) 2×10 N
9
(3) 2×10 N 5
(2) 1×10 N
(4) 3×10−5 N 9
(3) 2×10 N
(4) 3×10−5 N
9. Two tubes of radii r1 and r2, and lengths l1
and l2, respectively, are connected in series
and a liquid flows through each of them 9. ŒÊ ŸÁ‹ÿÊ° Á¡Ÿ∑§Ë ÁòÊíÿÊÿ¥ ∑˝§◊‡Ê— r1 ∞fl¥ r2 ÃÕÊ
in stream line conditions. P1 and P2 are ‹ê’Ê߸ÿÊ°, l1 fl l2 „Ò¥, ∑§Ê üÊáÊË ∑˝§◊ ◊¥ ¡Ê«∏Ê ªÿÊ „Ò
pressure differences across the two tubes.
•ÊÒ⁄U ©Ÿ◊¥ ∞∑§ Œ˝fl œÊ⁄UÊ ⁄UπËÿ ¬˝flÊ„ ◊¥ ’„ÃÊ „Ò–
l
If P2 is 4P1 and l2 is 1 , then the radius r2 ¬„‹Ë ÃÕÊ ŒÍ‚⁄UË Ÿ‹Ë ∑§ Á‚⁄UÊ¥ ∑§ ’Ëø ∑§ ŒÊ’ÊãÃ⁄U
4
∑˝§◊‡Ê— ÃÕÊ „Ò¥– ÿÁŒ ∑§Ê ◊ÊŸ ÃÕÊ
P1 P2 P2 4P1 l2
will be equal to :
l
∑§Ê ◊ÊŸ 1 „Ê ÃÊ ÁòÊíÿÊ ∑§Ê ◊ÊŸ „ÊªÊ —
(1) r1 r2
4
(2) 2r1
(3) 4r1 (1) r1
r
1 (2) 2r1
(4)
2
(3) 4r1
r
1
(4)
2
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(English+Hindi)
10. For the P-V diagram given for an ideal gas, 10. ∞∑§ •ÊŒ‡Ê¸ ªÒ‚ ∑§Ê P-V •Ê⁄Uπ ÁŒÿ ªÿ ÁøòÊ ◊¥
Œ‡ÊʸÿÊ ªÿÊ „Ò–
out of the following which one correctly
represents the T-P diagram ? ÁŒÿ ªÿ •Ê⁄UπÊ¥ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ‚„Ë T-P •Ê⁄Uπ
Œ‡ÊʸÿªÊ?
(1)
(1)
(2)
(2)
(3)
(3)
(4)
(4)
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