a uniform meter rule balances horizontally on a knife edge placed at 58cm mark when a weight of 20 gf is suspended from one end.What is the weight of the rule.

a uniform meter rule balances horizontally on a knife edge placed at 5

1.	a uniform meter rule balances horizontally on a knife edge placed at 58cm mark when a weight of 20 gf is suspended from one end.What is the weight of the rule.
Answer» Let, W be the mass of uniform scale, When the EDGE is placed at 58 cm MARK and 20 g weight suspended at one end, then uniform scale would be balaced. Applying here the principle of MOMENTS: The uniform scale weight would be balanced at MID point i.e. 50 cm of the scale. A weight of 20 g is placed at 58 cm mark. So the clockwise moment due to mass of the scale will be W(50-42) = 8W The anticlockwise moment due to mass 20G will be 20*(50-8) = 840. For equilibrium, clockwise moment = anticlockwise moment So, W x 8 = 840 W = 105 g The wight of the uniform meter = 105 g.

Discussion

No Comment Found

Related InterviewSolutions

1. Mind (ability, power) is one of the strongest and most useful powers youpossess2. Your mind mapping ability (declines, enhances) your brain functions.3. Lett brain controls all the muscles on the right-hand side of the body andthe (left, right) brain controls the left side. One hemisphere may be slightlydominant, as with left or right-handedness.139
8- A girl of mass 35 kg site on trolley of mas 5kg.initial velocity of trolley is im/s Come to rest after travelling 16m.find work done on trolley.come to meet after thavelling 16 m(a) work done ou trolley(b) work done by girl
Guys...this may seems to be hard...bt no...it's not...I need to know the solution of only part:-(d) & (e)(a)The moment of inertia of a disc of radius 0.5 m about its geometric axis is 2kg-m2. If a string is tied toits circumference and a force of 10 Newton is applied, the value of torque with respect to this axis will be(1) 2.5 N-m21275 N-m(3) 10 N-m(4) 20 N-m(b)In the above question, if the disc executes rotatory motion, its angular acceleration will be :-(1) 2.5 rad/sec(2) 5 rad/sec(3) 10 rad/sec(4) 20 rad/sec(c)In the above question, the value of its angular velocity after 2 seconds will be :-(1) 2.5 rad/sec(2) 5 rad/sec(3) 10 rad/sec(4) 20 rad/sec(d)In the above question, the change in angular momentum of disc in first 2 seconds in Nm second will be(1) 2.5(2) 5(3) 10(4) 20(e)In the above question, angular displacement of the disc, in first two second will be in radian:-(1) 2.5(2) 5(3) 10(4) 20Plz guys...I just don't wanna copy it in my hw...I wanna learn how the correct answer came out...so plz explain your solution with proper formula and sign and don't spam...plzWould be very thankful to you...!!!
Explain formation of rainbow with the help of a diagram.
consider the equation x=A sin (wt + pi/2) for a vibration object, where A is the amplitude. The Initial position of the object will be?
Table-1Articles/ Activities (A)Articles Materials used30-40 years ago B)clay jarsArticles Materials usedToday (C)Containers for storing picklePacking food for a journeyWater pipes in the houseplastic tiffin boxesCopper vesselsSteel vesselsVessels for cooking foodFuels used to cook foodFuel used in a train engineLuggage for carrying clothesWater buckets, pailser storages in housesMetalturksPlastic bucketsFurniture (chairs, cots)
What are your insights or refelction to the "fully essential to social and human developmet
acceleration of projection of a particle moving in a circle is related to the angular frequency w as?
4. Find the current of an electrical circuit that has a voltage supply of 20 Volts and a resistanceof 42.5. A motor with an operating resistance of 5 2 is connected to a voltage source. The current,in the circuit is 3 A. What is the voltage of the source?
The waveform of SHM is?

a uniform meter rule balances horizontally on a knife edge placed at 58cm mark when a weight of 20 gf is suspended from one end.What is the weight of the rule.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment