Self Studies

Work Energy and Power Test - 69

Result Self Studies

Work Energy and Power Test - 69
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    The graph below represents the relation between displacement $$x$$ and force $$F$$. The work done in displacing an object from $$x=8\ m$$ to $$x=16\ m$$  is approximately.

  • Question 2
    1 / -0
    What is the amount of work done in raising a glass of water weighing $$0.5 kg $$ through a height of $$50 cm ? (g = 10 m/s ^2 $$ )
    Solution

     

    The amount of the work done in raising the glass of the water through a height  is calculated by the amount of potential energy

    $$W = mgh$$

    By substituting the value in the above equation we get

    $$W = 0.5 \times 10 \times 0.5$$

    Hence the work done in raising the glass is$${\rm{2}}{\rm{.5J}}$$

  • Question 3
    1 / -0
    A mass $$m$$ is revolving in a vertical circle at the end of a string of length $$20\ cm$$. By how much times does the tension of the string at the lowest point exceed the tension at the topmost point-
    Solution
    Let tension $$T_1$$ at top most point is given by
    $$T_1 + mg = \dfrac{mV^2_1}{l}$$                ...(1)

     similarly, the tension $$T_2$$ at the lowest point is given by
    $$T_2 -mg = \dfrac{mV_2^2}{l}$$               ...(1)

     here, tension acting in words while balancing centrifugal  force  and weight of the body acting outwards
    $$T_2 - T_1 = \dfrac{mv^2_2}{l} +mg - \dfrac{mv_1^2}{l} +mg$$

    $$\dfrac{m}{l} (v^2_2 -V^2_1) + 2mg$$

     we know from kilometres
    $$v^2_1 = v^2_2 - 2g(2l)$$
    $$T_2 T_1 = \dfrac{m}{l} [2g(2l)] +2mg$$

    $$= 6mg$$
     Hence (C) is correct answer
  • Question 4
    1 / -0
    Two steel spheres approach each other head on with the same speed and collide elastically. After the collision one of the sphere's of radius r comes to rest, the radius of the other sphere is
    Solution

  • Question 5
    1 / -0
    An object of mass $$M_1$$ moving horizontally with speed u collides elastically with another object of mass $$M_2$$ at rest. Select correct statement.

    Solution

  • Question 6
    1 / -0
    As shown in the given figure the ball is given sufficient velocity at the lowest point to complete the circle. Length of string is $$1m$$. Find the tension in the string, when it is at $$60^{\circ}$$ with vertical position.
    (Mass of ball $$= 5\ kg$$).

    Solution

  • Question 7
    1 / -0
    A ball of radius r moving with a speed v collides elastically with another identical stationary ball. The impact parameter for the collision is b as shown in figure.

    Solution

  • Question 8
    1 / -0
    A stone of mass $$1kg$$ is tied to one end of a string of length $$0.5\ m$$. It is whirled in a veretical circular. If the maximum tension in the string is $$58.8N$$, the velocity at the top is
    Solution

  • Question 9
    1 / -0
    If increase in linear momentum of a body is 50%, then change in its kinetic energy is
    Solution

  • Question 10
    1 / -0
    In the figure shown initially spring is in uninstructed state & blocks are at rest. Now $$100N$$ force is appiled on block $$A$$ and $$B$$ as shown in figure. After some time velocity of $$'A'$$ becomes $$2m/s$$ and that of $$'B'\ 4m/s$$ and block $$A$$ displaced by amount $$10\ cm$$ and spring is streched by amount $$30\ cm$$. then find the work done by sprig force on $$A$$.

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Selfstudy
Selfstudy
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now