Self Studies
Selfstudy
Selfstudy

Electricity Test - 70

Result Self Studies

Electricity Test - 70
  • 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
    Choose the statement which is not correct in the case of an electric fuse.
    Solution
    In case of electric fuse, there is only a maximum limit on the current which can safely flow in the electric circuits. 
    There is a minimum limit on the current which can safely flow in the electric circuits.
  • Question 2
    1 / -0
    Variation of current and voltage in a conductor has been shown in the diagram below. The resistance of the conductor is.

    Solution
    from ohm's law $$R=\dfrac VI$$
    also, slope of given graph $$\dfrac VI$$
    so $$R=$$ slope of graph
    $$s=\dfrac{y_2-y_1}{x_2-x_1}$$

    $$s=\dfrac{4-1}{4-1}=1$$
    Resistance = 1 ohm
  • Question 3
    1 / -0
    Two wires of the same material are given. The first wire is twice as long as the second and has twice the diameter of the second. The resistance of the first will be.
    Solution
    we know $$R=\rho l /A$$
    so $$R\propto \dfrac{l}{r^2}$$
    According to the question:
    $$l_1=2l_2$$ and $$r_1=2r_2$$
    so $$R_1=\dfrac{2}{2^2}R_2=\dfrac{1}{2}R_2$$
    The resistance of the first will be Half of the second.
  • Question 4
    1 / -0
    Two resistors of resistance $$R_{1}$$ and $$R_{2}$$ having $$R_{1} > R_{2}$$ are connected in parallel. For equivalent resistance $$R$$ , the correct statement is
    Solution
    For parallel combination:
    $$\dfrac{1}{R_{Eq}}=$$$$\dfrac{1}{R_1}+$$$$\dfrac{1}{R_2}$$
    Equivalent resistance of parallel resistors is always less than any of the member of the resistance system.
    so $$R < R_{1}$$
  • Question 5
    1 / -0
    If three resistors of resistance $$2\Omega, 4\Omega$$ and $$5 \Omega$$ are connected in parallel then the total resistance of the combination will be
    Solution
    For parallel resistance connection we know that,
    $$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{19}{20}\Rightarrow R_{eq}=\dfrac{20}{19}\Omega$$
  • Question 6
    1 / -0
    Which of the following statement is false
    Solution
    According to Joule's heating law -  $$H = i^{2} Rt$$ 
    We also know that current $$i = \dfrac{q}{t}$$. 
    Hence $$H = \dfrac{q^{2}R}{t}$$ 
    $$\therefore H \propto q^{2}$$
  • Question 7
    1 / -0
    An electric heater kept in vacuum is heated continuously by passing electric current. Its temperature
    Solution
    Unless the heat generation rate is matched by the rate of radiation, the temperature will continue to rise.

    After that, the temperature will become constant.
  • Question 8
    1 / -0
    $$10$$ wires (same length, same area, same material) are connected in parallel and each has $$1\Omega$$ resistance, then the equivalent resistance will be
    Solution
    When $$n$$ no. of resistances are connected parallelly then

    $$ \dfrac{1}{R_{eq} } = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + ................$$

    $$ \dfrac{1}{R_{eq} } = \frac{n}{R}$$


    $$\Rightarrow$$ $$R_{eq} = \dfrac{R}{n}$$

    $$\Rightarrow$$ $$R_{eq}=\dfrac {R}{n}=\dfrac {1}{10}=0.1\Omega$$
  • Question 9
    1 / -0
    A solenoid is at potential of difference $$60\ V$$ and current flows through it is $$15$$ ampere, then the resistance of coil will be 
    Solution
    Given,
    Potential difference $$V=60\ V$$
    Current $$i=15\ A$$
    Resistance $$R=?$$

    Resistance $$=\dfrac {Potential\ difference}{Current}=\dfrac{60V}{15A}=4\Omega$$
  • Question 10
    1 / -0
    When 25 A current with 10 ohm resistance flowing per second then the power lost in the cable during transmission is
    Solution
    Power lost in cable $$= I^2R =10 \times (25)^{2} = 6250 W = 6.25 kW$$
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