The Complete Equilibrium
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- (M+m2)gtanθ
- (M+m2)gcotθ
- (M+m2)gcosθ
- (M+m2)gsinθ
- 2
- 3
- √2
- √3
- TA=2mg3; TB=mg3
- TA=mg3; TB=2mg3
- TA=TB=mg2
- TA=mg3, TB=mg4
- 30 N, 10 N
- 20 N, 20 N
- 30 N, 20 N
- 30 N, 30 N
- Net torque about points lying on axis of rotation must be zero.
- Net torque about points lying on the body must be zero
- Net torque about points lying outside the body must be zero
- All of these
- 5 kg
- 10 kg
- 8 kg
- 12 kg
Two forces with a magnitude $ 10$ units each are acting as shown in the diagram, their resultant magnitude is?
10 dynes 100 120° 40 dynes
- Torque of normal force on the block about centre is Mgasinθ2
- Torque of normal force on the block about the centre is Mgacosθ2
- Normal reaction force shifts by a distance atanθ2
- Normal reaction force shifts by a distance atanθ4
- True
- False
The magnitude of normal reaction exerted by the rod on the cylinder is
- mg4
- mg3
- mg2
- 2mg3
- 8.5 rad/s2
- 4.4 rad/s2
- 3.4 rad/s2
- 5.4 rad/s2
8 kg द्रव्यमान के एक गुटके को खुरदरे आनत तल पर रखा जाता है तथा इस पर चित्र में दर्शाए अनुसार तल के ऊपर की ओर निर्देशित 30 N का बल आरोपित किया जाता है। यदि तल का आनति कोण 60° है तथा गुटके व तल के मध्य स्थैतिक घर्षण गुणांक 0.4 है, तब गुटके द्वारा तल पर आरोपित कुल सम्पर्क बल किसके अनुदिश निर्देशित हो सकता है?
- −−→OA
- −−→OB
- −−→OC
- −−→OE
- 200 g
- 225 g
- 350 g
- 275 g
- N1=13mg, N2=23mg
- N1=23mg, N2=13mg
- N1=mg2, N2=mg2
- N1=mg4, N2=3mg4
- 1.5 m away from fulcrum
- 1.2 m away from fulcrum
- 1.7 m away from fulcrum
- None of these.
- 5 kg
- 10 kg
- 8 kg
- 12 kg
A 1kg rod of length 1m is pivoted at its centre and two masses of 5kg and 2kg and are hung from the ends as shown in figure. Find the tension in the supports to the blocks of mass 2kg and 5kg (g=9.8ms2)
20.6N, 29N
20.6N, 20.6N
27.6N, 20.6N
27.6N, 29N
The magnitude of normal reaction exerted by the rod on the cylinder is
- mg4
- mg3
- mg2
- 2mg3
- Magnitude of normal reaction by wall on ladder at point B will increase.
- Magnitude of normal reaction by wall on ladder at point B will decrease.
- Magnitude of normal reaction by floor on ladder at point A will remain unchanged.
- Magnitude of frictional force by the floor on ladder at point A will increase.
- 2
- √3
- √2
- 1√3
- Only translational equilibrium
- Only rotational equilibrium
- Rotational and translational equilibrium
- Neither rotational nor translational equilibrium
- μ1=0μ2≠0 and N2 tan θ=mg/2
- μ1≠0μ2=0 and N1 tan θ=mg/2
- μ1≠0μ2≠0 and N2=mg1+μ1μ2
- μ1=0μ2≠0 and N1 tan θ=mg/2
- 8.5 rad/s2
- 4.4 rad/s2
- 3.4 rad/s2
- 5.4 rad/s2
- X=3 N, Y=3 N
- X=0, Y=6 N
- X=6 N, Y=3 N
- X=6 N, Y=6 N
In the system shown in figure blocks A and B have mass m1=2kg and m2=267 kg respectively. Pulley having moment of inertia I = 0.11 kg m2 can rotate without friction about a fixed axis. Inner and outer radii of pulley are a = 10 cm and b = 15 cm respectively. B is hanging with the thread wrapped around the pulley, while A lies on a rough inclined plane.
Coefficient of friction being μ=√310
ColumnIColumnIIP.Tension in the thread connecting block Aw.26NQ.Tension in the thread connecting block Bx.2ms2R accelerationofAR.Acceleration of Ay.3ms2S.Acceleration of Bz.17N
P-w, Q-z, R-x, S-y
P-z, Q-w, R-y, S-x
P-z, Q-w, R-x, S-y
P-z, Q-z, R-y, S-y
The ratio of magnitude of frictional force on the cylinder due to the rod and the magnitude of frictional force on the cylinder due to the inclined plane is:
- 1:1
- 2:√3
- 2:1
- √2:1
- mg8
- 2mg7
- 3mg5
- 3mg8
- slides down
- topples over
- remains in complete equilibrium
- cannot be determined.
The ratio of magnitude of frictional force on the cylinder due to the rod and the magnitude of frictional force on the cylinder due to the inclined plane is:
- 1:1
- 2:√3
- 2:1
- √2:1