21 NEET previous-year questions on Motion in a Plane, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
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3.5
5.9
16.3
110.8
Solution
The range of a projectile is given by . For identical trajectories, the ranges are equal, so . Therefore, option (a) is correct.
t= π 4ω
t= π 2ω
t= π ω
t=0
Solution
A⃗⃗ =coswt î+sinwt ĵ B⃗⃗ =coswt 2 î+sinwt 2 ĵ for A⃗⃗ .B⃗⃗ =0 A⃗⃗ .B⃗⃗ =0=cos wt.coswt 2 +sinwt.sinwt 2 =cos(wt−wt 2)=cos( wt 2) So wt 2 = π 2 ⇒ t= π w
Acceleration vector is along −R⃗⃗ .
Magnitude of acceleration vector is v2 R , where v is the velocity of particle.
Magnitude of the velocity of particle is 8 meter/second
Patch of the particle is a circle of radius 4 meter.
Solution
www.vedantu.com 33 x= 45 m 2πt, y=4c os(2πt) Squiring and adding ⇒ Circular motion V=ω= (2π)(4)=8π
r⃗ 1−r⃗ 2 |r⃗ 1−r⃗ 2|= v⃗⃗ 2−v⃗⃗ 1 |v⃗⃗ 2−v⃗⃗ 1|
r 1∙v⃗ 1=r 2∙v⃗ 2
r 1×v⃗ 1=r 2×v⃗ 2
r 1−r 2=v⃗ 1−v⃗ 2
Solution
For two particles to collide, the direction of the relative velocity of one with respect to other should be directed towards the relative position of the other particle i.e. r⃗ 1−r⃗ 2 |r⃗ 1−r⃗ 2| → direction of relative position of 1 w.r.t..2. & v⃗⃗ 2−v⃗⃗ 1 |v⃗⃗ 2|−v⃗⃗ 1 → direction of velocity of 2 w.r.t.1 www.vedantu.com 32 So for collision of A & B r 1−r 2 |r 1−r 2|= v⃗ 2−v⃗ 1 |v⃗ 2−v⃗ 1|
0.1 m/s2
0.15 m/s2
0.18 m/ s2
0.2m s2
Solution
Tangential acceleration at = rα = constant = K α = K r At the end of second revoluation angular velocity is w then w2 − w0 2 = 2 ∝ θ w2 − O2 = 2(K r)(4π) w2 = 8πK r K.E. of the particle is = K.E.= 1 2mv2 K.E.= 1 2mr2w2 K.E.= 1 2 m(r2)(8πK r ) 8 ×10−4 = 1 2 × 10× 10−3 × 6.4 ×10−2 × 3.14× K 𝐾 = 2 6.4 × 3.14= 0.1 𝑚 𝑠𝑠𝑒𝑐2
Velocity and acceleration both are perpendicular to r .
Velocity and acceleration both are parallel to r .
Velocity is perpendicular to r and acceleration is directed towards the origin.
Velocity is perpendicular to r and acceleration is directed away from the origin.
Solution
Position vector is r = cosω + x ̂ +sinω + y ̂ Velocity of particle is v⃗ = dr⃗ dt v⃗ = sinωt.ωx ̂ + cosωt.ωy ̂ v⃗ = ω(−sinωtx ̂ + cosωty ̂) Acceleration of the particle is a⃗ = dv⃗ dt a⃗ = −ω2(cosωtx ̂ +sinωty ̂) a⃗ = −ω2r , So direction of r and a⃗ are opposite. v⃗ .a⃗ = 0 ⇒ v⃗ ⊥ a⃗ v⃗ .r = 0 ⇒ v⃗ ⊥ r So, ans is (Velocity is perpendicular to r and acceleration is directed towards the origin.)
0𝑜
90𝑜
45𝑜
180𝑜
Solution
www.vedantu.com 24 |A⃗⃗ + B⃗⃗ | = |A⃗⃗ − B⃗⃗ | A2 + B2 + 2ABcosθ = A2 + B2 = 2ABcosθ 4ABcosθ = 0 cosθ = 0 θ =90o
√gR2 μs+tanθ 1−μs tanθ
√gR μs+tanθ 1−μs tanθ
√ g R μs+tanθ 1−μstanθ
√ g R2 μs+tanθ 1−μstanθ
Solution
www.vedantu.com 28 N =mgcosθ +mv2 r sinθ fmax = μmgcosθ +μmv2 r sinθ mgsinθ + μmgcosθ +μmv2 r sinθ =mv2 r cosθ gsinθ + gcosθ =V2 r (cosθ − μsinθ) gr[ tanθ + μ 1 + μtanθ] = V2 www.vedantu.com 29 CHEMISTRY
T
2mvT l+
2mvT l-
Zero
Solution
Centripetal force 2mv l ⎛⎞ ⎜⎟⎜⎟⎝⎠ is provided by tension so the net force will be equal to tension i.e., T.
inclined at an angle of 60° from vertical
the mass is at the highest point
the wire is horizontal
the mass is at the lowest point
Solution
Tension is maximum at bottom ⇒ Mass is at the lowest point
1 : 4
2 : 1
1: 2
4 : 1
Solution
𝑅= 𝑚𝑣 𝑞𝐵 𝑅𝐻 𝑅∝ = 𝑚𝑣 𝑒𝐵 ×2𝑒𝐵 𝑚𝑣=2∶1
electric heater
induction furnace
magnetic braking in train
electromagnet
Solution
Conceptual.
3 : 1
2 : 1
1 : 2
2 : 3
Solution
1 𝐹1 = 1 𝑓+ 1 𝑓= 2 𝑓 𝐹1=𝑓/2 1 𝐹2 = 1 𝑓− 1 𝑓+1 𝑓= 1 𝑓 𝐹2=𝑓 𝐹1∶ 𝐹2=1 ∶2
45° west
30° west
0°
60° west
Solution
2 1 20 10 mv Rvcos −= − =−= θ θ =120° ∴ angle with north = 30° west
1 2 2 1 2 gT cos R − θ= π
1 2 2 1 2 R cos gT − π θ=
1 2 2 1 2 R sin gT − π θ=
1 2 2 1 2 2gT sin R − θ= π
Solution
The maximum height for a projectile is given by . For a particle moving in a circle with radius and period , the speed . Given , substituting gives . Taking the inverse sine, . Simplifying, . However, the correct form is , which matches option (b).
2π
4π
12π
104π
Solution
Angular acceleration ω −ωα= fi t 23120 rad/s60 πω= ×f 21200 rad/s60 πω= ×i ⇒ (3120 1200) 2 416 60 −πα= × = π
Zero
15 3 ms −
5 ms–1
10 ms–1
Solution
At highest point vertical component of velocity become zero. At highest point speed of object = 10cos30° 5 3 m/s= - 16 - NEET (UG)-2022 (Code-Q1)
along the axis of rotation
along the radius, away from centre
along the radius towards the centre
along the tangent to its position
Solution
Angular acceleration is a vector quantity directed along the axis of rotation, perpendicular to the plane of motion. This aligns with the definition in NCERT XI chapter Motion in a Plane, so option (a) is correct.
3000 m
2800 m
2000 m
1000 m
Solution
The maximum height is given by . Substituting, , so option (d) is correct.
Constant velocity
Constant acceleration
Constant velocity but varying acceleration
Varying velocity and varying acceleration
Solution
A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously.
T
4T
4 T
2T
Solution
2=ω Tm 2(2 )′ = ω Tm 4′=TT SECTION-B
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