12th Grade > Physics
MOTION IN TWO DIMENSION MCQs
Total Questions : 31
| Page 3 of 4 pages
Answer: Option B. -> 4 rad/sec
:
B
Centripental force = breaking force
⇒ m ω2 r = Breaking stress × cross sectional area
⇒ m ω2 r = p × A ⇒ ω = √p×Amr = √4.8×107×10−610×0.3
∴ ω = 4 rad/sec
:
B
Centripental force = breaking force
⇒ m ω2 r = Breaking stress × cross sectional area
⇒ m ω2 r = p × A ⇒ ω = √p×Amr = √4.8×107×10−610×0.3
∴ ω = 4 rad/sec
Answer: Option A. -> 2 mg
:
A
When body is released from the position p (inclined at angle θ from vertical) then velocity at mean position
v = √2gl(1−cosθ)
∴ Tension at the lowest point = mg + mv2l
= mg + ml[2gl(1 - cos 60)] = mg + mg = 2mg
:
A
When body is released from the position p (inclined at angle θ from vertical) then velocity at mean position
v = √2gl(1−cosθ)
∴ Tension at the lowest point = mg + mv2l
= mg + ml[2gl(1 - cos 60)] = mg + mg = 2mg
Answer: Option A. -> It hits the ground at a horizontal distance 1.6 m from the edge of the table
:
A
Vertical component of velocity of ball at point P
vv = 0 + gt = 10 × 0.4 = 4 m/s
Horizontal component of velocity = inital velocity
⇒ vH = 4 m/s
So the speed with which it hits the ground
v = √(vH2+vy2) = 4√2 m/s
and tan θ= vVvH =44 = 1 ⇒ θ = 45∘
It means the ball hits the ground at an angle of 45∘ to the horizontal.
Height of the table h = 12 g t2 = 12 × 10 × (0.4)2 = 0.8m
Horizontal distance travelled by the ball from the edge of table h = ut = 4 × 0.4 = 1.6m
:
A
Vertical component of velocity of ball at point P
vv = 0 + gt = 10 × 0.4 = 4 m/s
Horizontal component of velocity = inital velocity
⇒ vH = 4 m/s
So the speed with which it hits the ground
v = √(vH2+vy2) = 4√2 m/s
and tan θ= vVvH =44 = 1 ⇒ θ = 45∘
It means the ball hits the ground at an angle of 45∘ to the horizontal.
Height of the table h = 12 g t2 = 12 × 10 × (0.4)2 = 0.8m
Horizontal distance travelled by the ball from the edge of table h = ut = 4 × 0.4 = 1.6m
Answer: Option B. -> 12
:
B
By using equation ω2 = ω02 - 2 αθ
ω022 = ω02 - 2α(2πn) ⇒ α = 34 ω024π×36 , (n = 36) ..(ii)
Now let fan completes total n′ revolution from the starting to come to rest
0 = ω02 - 2α(2πn) ⇒ n′ = ω024απ
substituting the value of α from equation (i)
n′ = ω024π 4×4π×363ω02 = 48 revolution
Number of rotation = 48 - 36 = 12
:
B
By using equation ω2 = ω02 - 2 αθ
ω022 = ω02 - 2α(2πn) ⇒ α = 34 ω024π×36 , (n = 36) ..(ii)
Now let fan completes total n′ revolution from the starting to come to rest
0 = ω02 - 2α(2πn) ⇒ n′ = ω024απ
substituting the value of α from equation (i)
n′ = ω024π 4×4π×363ω02 = 48 revolution
Number of rotation = 48 - 36 = 12
Answer: Option A. -> √98 m/s
:
A
In this problem it is assuemed that particle although moving in a vertical loop but its speed remain constant.
Tenstion at lowest point Tmax = mv2r + mg
Tension at highest point Tmin = mv2r - mg
TmaxTmin = mv2r+mgmv2r−mg = 53
by solving we get , v=√4gr = √4×9.8×2.5 = √98 m/s
:
A
In this problem it is assuemed that particle although moving in a vertical loop but its speed remain constant.
Tenstion at lowest point Tmax = mv2r + mg
Tension at highest point Tmin = mv2r - mg
TmaxTmin = mv2r+mgmv2r−mg = 53
by solving we get , v=√4gr = √4×9.8×2.5 = √98 m/s
Answer: Option A. -> 172 m/s
:
A
The maximum velocity for a baked road with friction,
v2 = gr(μ+tanθ1−μtanθ)
⇒ v2 = 9.8 × 1000 × (0.5+11−0.5×1) ⇒ v = 172 m/s
:
A
The maximum velocity for a baked road with friction,
v2 = gr(μ+tanθ1−μtanθ)
⇒ v2 = 9.8 × 1000 × (0.5+11−0.5×1) ⇒ v = 172 m/s