Self Studies

Units and Measurements Test - 85

Result Self Studies

Units and Measurements Test - 85
  • 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
    A particle of mass mm collides with a stationary particle and continues to move at an angle of 45o{45}^{o} with respect to the original direction. The second particle also recoils at an angle of 45o{45}^{o} to this direction. The mass of the second particle is (collision is elastic)
    Solution

  • Question 2
    1 / -0

    Two resistors of resistances R1=150±2 ΩR_1= 150\pm2 \space \Omega and R2=220±6  ΩR_2 = 220\pm6 \space \Omega areconnected in parallel combination.Calculate the equivalent resistance.

    Solution
     Given R1=150±2ΩR2=220±6Ω \text { Given }-R_{1}=150 \pm 2 \Omega \quad R_{2}=220 \pm 6 \Omega  
     When connect in Parallel we have - Req=R1R2R1+R2Req=150220150+220=89.19 s \begin{array}{l} \text { When connect in Parallel we have - } \\ \qquad R_{e q}=\frac{R_{1} \cdot R_{2}}{R_{1}+R_{2}} \Rightarrow R_{e q}=\frac{150 \cdot 220}{150+220}=89.19 \mathrm{~s} \end{array}
      for calculating error  we have 1Req=1R1+1R2 \begin{array}{l} \text { for calculating error }- \\ \text { we have } \frac{1}{\operatorname{Req}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \end{array}  
     differentiate both Sides d(Req)(Req)2=dR1(R1)2dR2(R2)2 \begin{array}{l} \text { differentiate both Sides } \Rightarrow \\ \qquad-\frac{d\left(R_{e q}\right)}{\left(R_{e q}\right)^{2}}=-\frac{d R_{1}}{\left(R_{1}\right)^{2}}-\frac{d R_{2}}{\left(R_{2}\right)^{2}} \end{array}  
    d( Req)=(89.19)2[2(150)2+6(220)2] \Rightarrow d(\text { R}_{eq})=(89.19)^{2} \cdot\left[\frac{2}{(150)^{2}}+\frac{6}{(220)^{2}}\right]  
    d(Req)=1.69Ω Now we can write Req=89.19±1.69Ω90Ω can be be the maximum suitable answer.  \begin{array}{l} d\left(R_{e q}\right)=1.69 \Omega \\ \text { Now we can write } R_{e q}=89.19 \pm 1.69 \Omega \\\Rightarrow \text{90}\Omega \text { can be be the maximum suitable answer. } \end{array}
  • Question 3
    1 / -0
    Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1R_1 and R2R_2 respectively. The ratio of the mass of X to that of Y is?
  • Question 4
    1 / -0
    A body is dropped from a height h. If it acquired a momentum p, Then the mass of the body is 
  • Question 5
    1 / -0
    A body of mass m1m_1, moving with a uniform velocity of 50ms1m{s^{ - 1}} collides with another body of mass m2m_2 at rest and then two together start moving with a velocity 40ms1m{s^{ - 1}} . The ratio of their masses (m1m2)\left( {\frac{{{m_1}}}{{{m_2}}}} \right) is 
    Solution

  • Question 6
    1 / -0
    Two particle X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1R_1 and R2R_2 respectively. The ratio of the mass X to the Y is:
    Solution

  • Question 7
    1 / -0
    Which of the following quantity is dimension less?
    Solution

  • Question 8
    1 / -0
    van der Waal's constant a'a' has the dimensions of 
    Solution

  • Question 9
    1 / -0
    The linear mass density i.e. mass per unit length of a rod of length L is given by ρ\rho = ρ0\rho_{0}(1 + xL\dfrac{x}{L}), where ρ0\rho_{0} is constant , x is distance from the left end. Find  the c.o.m from the left end.

  • Question 10
    1 / -0
    Two isotopes of an element XX are present in the ratio of 1:21:2, having mass number MM and (M+0.5M+0.5) respectively. Find the mean mass number of XX
    Solution

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
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