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Physics Test-1

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Physics Test-1
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  • Question 1
    4 / -1

    The IUPAC name for \(\mathrm{CH}_{3} \mathrm{CH}=\underset{\overset{|}NH_2}{\mathrm{CHCH}_{2} \mathrm{CHCH}_{2} \mathrm{COOH}}\) is:

    Solution

    As - COOH group is the highest priority group, it is numbered one. So, the IUPAC name is \(3-\)amino\(-5-\)heptanoic acid.

    \({\overset{7}{\mathrm{C}}H}_{3}-{\overset{6}{\mathrm{C}}H}={\overset{5}{\mathrm{CH}}}-{\overset{4}{\mathrm{CH}_{2}}}-{\underset{\mathrm{NH_2}}{\underset{|}{{\overset{3}{\mathrm{CH}}}}}}-{\overset{2}{\mathrm{CH}_{2}}}-{\overset{1}{\mathrm{CH}}} \mathrm{OO}\)

  • Question 2
    4 / -1

    Which of the following is a thermodynamics law?

    Solution

    Thermodynamics is primarily based on a set of four rules that are universally applicable when applied to systems that fall within their respective limitations. They are as follows:

    • Zeroth law of thermodynamics
    • First law of thermodynamics
    • Second law of thermodynamics
    • Third law of thermodynamics
  • Question 3
    4 / -1

    A moving coil galvanometer can be converted into a ammeter by connecting:

    Solution

    A moving coil galvanometer can be converted into a ammeter by connecting a low resistance in parallel to the moving coil galvanometer.

    A galvanometer can be converted into an ammeter by connecting a shunt resistance in parallel to it. The shunt resistance should have very low resistance. So, the ammeter (the parallel combination of galvanometer and shunt resistance) will have low resistance.

  • Question 4
    4 / -1

    A circuit has a section \({AB}\) as shown in figure. The emf of the source equals \({E}=10 {~V}\), the capacitor capacitances are equal to \({C}_{1}=1.0 \mu {F}\) and \({C}_{2}=2.0 \mu {F}\) and the potential difference \({V}_{{A}}-{V}_{{B}}=5.0 {~V}\). Find the voltage across each capacitor.

    Solution

    Let the charge distribution be as shown in figure:


    \(\therefore V_{A}-V_{B}=\frac{q}{C_{1}}-E+\frac{q}{C_{2}}\)

    or, \( \left(V_{A}-V_{B}\right)+E=q\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right)\)

    \( \left(V_{A}-V_{B}\right)+E=\frac{{q}\left({C}_{2}+C_{1} \right)}{{C}_{1} {C}_{2}}\)

    \(\therefore  {q}=\frac{\left[\left({V}_{{A}}-{V}_{{B}}\right)+{E}\right] {C}_{1} {C}_{2}}{{C}_{1}+{C}_{2}}\)

    Voltage across \(C_{1}\) is \(V_{1}=\frac{q}{C_{1}}\)

    \(=\frac{ \left [\left(V_{A}-V_{B}\right)+E\right] C_{2}}{C_{1}+C_{2}}\) 

    \(=\frac{(5+10) 2.0}{1.0+2.0}\)

    \(=10\) Volt

    Voltage across \(C_{2}\) is \(V_{2}=\frac{q}{C_{2}}\)

    \(=\frac{ \left [\left(V_{A}-V_{B}\right)+E\right] C_{1}}{C_{1}+C_{2}}\) 

    \(=\frac{(5+10) 1.0}{1.0 \times 2.0}\)

    \(=5\) Volt

  • Question 5
    4 / -1

    The potential energy of a particle varies with distance \(x\) from a fixed origin as \(U=\frac{A \sqrt{x}}{x^{2}+B}\) where \(\mathrm{A}\) and \(\mathrm{B}\) are constants. The dimensions of \(\mathrm{A}\) are:

    Solution

    Given,

    \(U=\frac{A \sqrt{x}}{x^{2}+B}\).....(i)

    Where \(U =\) Potential energy

    \(x=\) distance

    Dimensional formula of potential energy,

    \(U=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]\)

    Dimensional formula of distance,

    \(x=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\right]\)

    From equation (i), we can write

    \(\left[x^{2}\right]=[B]\)

    \([B]=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\right]^{2}\)

    \([B]=\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\)

    Dimensional formula of \(B\) \(=\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\)

    We can write equation (i) as,

    \(U=\frac{A \sqrt{x}}{x^{2}}\)

    \(U=\frac{A}{x^{\frac{3}{2}}}\)

    \(A=U x^{\frac{3}{2}}\)

    \(A=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right] \times\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\right]^{\frac{3}{2}}\)

    \(A=\left[\mathrm{M}^{1} \mathrm{~L}^{\frac{7}{2}} \mathrm{~T}^{-2}\right]\)

  • Question 6
    4 / -1

    The thermal capacity of a body depends on the:

    Solution
    • The heat capacity of a body is the quantity of heat required to raise the temperature of the body by one degree.
    • It is measured in J/K and depends on the mass.
    • The specific heat capacity is designed to eliminate mass as a factor by dividing by mass, so it depends only on the material the object is made of.
    • It is measured in units of J/kg K.
  • Question 7
    4 / -1

    The radius of first orbit of hydrogen atom is \(0.53~ A^{\circ}\). The radius of its fourth orbit will be

    Solution

    Given: Radius of first orbit \(=0.53 ~A^{\circ}\)

    We know that:

    Radius of the orbit of \(e^{-}( r )\).

    In this condition, radius is directly proportional to number of orbits.

    \(r \propto n^{2}, ~n \rightarrow n ^{\text {th }}\) orbit

    Let \(r_2\) be the radius of \(4^{\text {th }}\) orbit.

    \(\Rightarrow \frac{r_1}{r_2}=\frac{n_1^2}{n_2^2} \).

    \(\Rightarrow r _1=\frac{ n _1^2}{ n _2^2} \cdot r _2\)

    For \(4^{\text {th }}\) orbit, \(n _2=4\) and \(n _1=1\)

    \(\Rightarrow r _2=(4)^2 \cdot r _1=16 \times 0.53 ~A ^{\circ}\)

    \(\Rightarrow r _2=8.48 ~A ^{\circ}\)

  • Question 8
    4 / -1

    The combination of gates shown in the circuit is equivalent to :

    Solution

    OR Gate

    An OR gate is a digital logic gate with two or more inputs and one output that performs logical disjunction. The output of an OR gate is true when one or more of its inputs are true. If all of an OR gate's inputs are false, then the output of the OR gate is false.

  • Question 9
    4 / -1

    Kepler's second law (law of areas) is nothing but a statement of:

    Solution

    Kepler's second law (law of areas) is nothing but a statement of conservation of angular momentum.

    Kepler’s laws of planetary motion:

    • The law of Orbits: Every planet moves around the sun in an elliptical orbit with the sun at one of the foci.
    • The law of Area: The line joining the sun to the planet sweeps out equal areas in equal interval of time. i.e. areal velocity is constant.
    • According to this law, the planet will move slowly when it is farthest from the sun and more rapidly when it is nearest to the sun. It is similar to the law of conservation of angular momentum.
    • The law of periods: The square of the period of revolution (T) of any planet around the sun is directly proportional to the cube of the semi-major axis of the orbit i.e. T2 ∝ r3
  • Question 10
    4 / -1

    Which of the following fertilizers has the highest nitrogen percentage?

    Solution

    Urea has the highest nitrogen percentage i.e.,\((46.6\%)\).

    In other compounds are,

    Ammonium sulphate \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}=21.2 \%\)

    Calcium cyanamide \(\mathrm{CaCN}_{2}=35.0 \%\) 

    And Ammonium nitrate \(\mathrm{NH}_{4} \mathrm{NO}_{3}=35.0 \%\)

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