VITEEE 2016 :: Physics :: General questions

• Physics - General questions

A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is (Take g= 10 ms^-2)

Answer:

Explanation:

$a=\mu g=\frac{6}{10}$     [using v= u+at]

$\Rightarrow\mu =\frac{6}{10\times g}=\frac{6}{10\times10}=0.06$

A hydrogen atom is an exciting state of principal quantum number (n), it emits a photon of wavelength ($\lambda$). when it returns to the ground state. the value of n is 

Answer:

Explanation:

As   $\frac{1}{\lambda}= R\left(\frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}}\right)$

 $\therefore$    $\frac{1}{\lambda}= R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}_{}}\right)$

 Mutliply both sides by $\lambda$

 $1= \lambda R\left(1-\frac{1}{n^{2}_{}}\right)or \frac{1}{\lambda R}= 1-\frac{1}{n^{2}}$

 or   $ \frac{1}{n^{2}}=1-\frac{1}{\lambda R}=\frac{\lambda R-1}{\lambda R}$

   or    $n= \sqrt{\frac{\lambda R}{\lambda R-1}}$

 If the nuclear radius of ²⁷ Al is 3.6 Fermi, the approximate nuclear radius of ⁶⁴Cu  in Fermi is

Answer:

Explanation:

 Nuclear radius , $r\propto A^{1/3}$

 Where A is mass number

 $\frac{r_{Cu}}{r_{Al}}=\left(\frac{A_{Cu}}{A_{Al}}\right)^{1/3}=\left(\frac{64}{27}\right)^{1/3}$

$\Rightarrow r_{Cu}=3.6\left(\frac{4}{3}\right)=4.8 Fermi$

 The half-life of radioactive Radon is 3.8 days. the time at the end of which 1/20 th of the radon sample will remain undecayed is (given  $\log_{10}e=0.43434 $)

Answer:

Explanation:

t_1/2  =3.8 day

 $\therefore$    $\lambda= \frac{0.693}{t_{1/2}}=\frac{0.693}{3.8}=0.182$

 If the initial number of atom is a = A₀ then after time r the number of atoms is a/20= A , we have to find t.

 $t= \frac{2.303}{\lambda}\log\frac{A_{0}}{A}=\frac{2.303}{0.182}\log\frac{a}{a/20}$

  $=\frac{2.303}{0.182}\log20=16.46$  days

 An unpolarized beam of intensity  I₀ is incident on a pair of nicols making an angle of 60° with each other. The intensity  of light emerging from the pair is 

Answer:

Explanation:

According to Malu's law  $I= I_{0}\cos^{2}\theta=I_{0}(\cos^{2}60^{0})$

 = $ I_{0}\times(\frac{1}{2})^{2}= \frac{I_{0}}{4}$

• Physics - General questions