NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
TEST SERIES (PHYSICS FST)
266260
If the momentum of electron is changes by \(P\), then the de-Broglie wavelength associated with it changes by \(0.5 \%\). The initial momentum of electron will be:
266261
A car moves with uniformacceleration up to some distance. Initial and final velocities are \(u\) and \(v\) then velocity at half way of the car will be:
1 \(\frac{1}{2}(u+v)\)
2 \(\frac{1}{2}\left(u^2+v^2\right)\)
3 \(\sqrt{\frac{1}{2}\left(u^2+v^2\right)}\)
4 \(\frac{1}{2} \sqrt{\left(u^2+v^2\right)}\)
Explanation:
c Suppose velocity at mid point is V . \[ \begin{aligned} & y^2=u^2+2 a s \\ & y^2=y^2+2 a s \\ & y^2-v^2=u^2-v^2 \\ & 2 v^2=u^2+v^2 \\ & y^2=\frac{1}{2}\left(u^2+v^2\right) \\ & V=\sqrt{\frac{1}{2}\left(u^2+v^2\right)} \end{aligned} \]
**NCERT-XI-I-18**
TEST SERIES (PHYSICS FST)
266262
lonisation potential of hydrogen atom is 13.6 eV . The energy required to remove an electron from the second orbit of hydrogen is (in eV ):
1 3.4
2 1.89
3 10.2
4 8.5
Explanation:
a Energy of electron in \(n^{\text {th }}\) state of hydrogen atom \(E=\frac{-13.6}{\pi^2} e V \) Energy of electron in second orbit \(E_2=\frac{-13.6}{(2)^2}=-3.4 \mathrm{eV} \) Thus 3.4 eVenergyis required to remove an electron from second orbit
**NCERT-XII-II-300**
TEST SERIES (PHYSICS FST)
266263
A p-n photodiode is fabricated from a semiconductor with a band gap of 3.1 eV . It can detect a signal of wavelength :
1 4000 nm
2 6000 nm
3 4000 A
4 6000 A
Explanation:
c \(\mathrm{A}-\frac{12400 \mathrm{eV}}{3.1 \mathrm{eV}}\) A \(=4000 \mathrm{~A}\).
266260
If the momentum of electron is changes by \(P\), then the de-Broglie wavelength associated with it changes by \(0.5 \%\). The initial momentum of electron will be:
266261
A car moves with uniformacceleration up to some distance. Initial and final velocities are \(u\) and \(v\) then velocity at half way of the car will be:
1 \(\frac{1}{2}(u+v)\)
2 \(\frac{1}{2}\left(u^2+v^2\right)\)
3 \(\sqrt{\frac{1}{2}\left(u^2+v^2\right)}\)
4 \(\frac{1}{2} \sqrt{\left(u^2+v^2\right)}\)
Explanation:
c Suppose velocity at mid point is V . \[ \begin{aligned} & y^2=u^2+2 a s \\ & y^2=y^2+2 a s \\ & y^2-v^2=u^2-v^2 \\ & 2 v^2=u^2+v^2 \\ & y^2=\frac{1}{2}\left(u^2+v^2\right) \\ & V=\sqrt{\frac{1}{2}\left(u^2+v^2\right)} \end{aligned} \]
**NCERT-XI-I-18**
TEST SERIES (PHYSICS FST)
266262
lonisation potential of hydrogen atom is 13.6 eV . The energy required to remove an electron from the second orbit of hydrogen is (in eV ):
1 3.4
2 1.89
3 10.2
4 8.5
Explanation:
a Energy of electron in \(n^{\text {th }}\) state of hydrogen atom \(E=\frac{-13.6}{\pi^2} e V \) Energy of electron in second orbit \(E_2=\frac{-13.6}{(2)^2}=-3.4 \mathrm{eV} \) Thus 3.4 eVenergyis required to remove an electron from second orbit
**NCERT-XII-II-300**
TEST SERIES (PHYSICS FST)
266263
A p-n photodiode is fabricated from a semiconductor with a band gap of 3.1 eV . It can detect a signal of wavelength :
1 4000 nm
2 6000 nm
3 4000 A
4 6000 A
Explanation:
c \(\mathrm{A}-\frac{12400 \mathrm{eV}}{3.1 \mathrm{eV}}\) A \(=4000 \mathrm{~A}\).
266260
If the momentum of electron is changes by \(P\), then the de-Broglie wavelength associated with it changes by \(0.5 \%\). The initial momentum of electron will be:
266261
A car moves with uniformacceleration up to some distance. Initial and final velocities are \(u\) and \(v\) then velocity at half way of the car will be:
1 \(\frac{1}{2}(u+v)\)
2 \(\frac{1}{2}\left(u^2+v^2\right)\)
3 \(\sqrt{\frac{1}{2}\left(u^2+v^2\right)}\)
4 \(\frac{1}{2} \sqrt{\left(u^2+v^2\right)}\)
Explanation:
c Suppose velocity at mid point is V . \[ \begin{aligned} & y^2=u^2+2 a s \\ & y^2=y^2+2 a s \\ & y^2-v^2=u^2-v^2 \\ & 2 v^2=u^2+v^2 \\ & y^2=\frac{1}{2}\left(u^2+v^2\right) \\ & V=\sqrt{\frac{1}{2}\left(u^2+v^2\right)} \end{aligned} \]
**NCERT-XI-I-18**
TEST SERIES (PHYSICS FST)
266262
lonisation potential of hydrogen atom is 13.6 eV . The energy required to remove an electron from the second orbit of hydrogen is (in eV ):
1 3.4
2 1.89
3 10.2
4 8.5
Explanation:
a Energy of electron in \(n^{\text {th }}\) state of hydrogen atom \(E=\frac{-13.6}{\pi^2} e V \) Energy of electron in second orbit \(E_2=\frac{-13.6}{(2)^2}=-3.4 \mathrm{eV} \) Thus 3.4 eVenergyis required to remove an electron from second orbit
**NCERT-XII-II-300**
TEST SERIES (PHYSICS FST)
266263
A p-n photodiode is fabricated from a semiconductor with a band gap of 3.1 eV . It can detect a signal of wavelength :
1 4000 nm
2 6000 nm
3 4000 A
4 6000 A
Explanation:
c \(\mathrm{A}-\frac{12400 \mathrm{eV}}{3.1 \mathrm{eV}}\) A \(=4000 \mathrm{~A}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
TEST SERIES (PHYSICS FST)
266260
If the momentum of electron is changes by \(P\), then the de-Broglie wavelength associated with it changes by \(0.5 \%\). The initial momentum of electron will be:
266261
A car moves with uniformacceleration up to some distance. Initial and final velocities are \(u\) and \(v\) then velocity at half way of the car will be:
1 \(\frac{1}{2}(u+v)\)
2 \(\frac{1}{2}\left(u^2+v^2\right)\)
3 \(\sqrt{\frac{1}{2}\left(u^2+v^2\right)}\)
4 \(\frac{1}{2} \sqrt{\left(u^2+v^2\right)}\)
Explanation:
c Suppose velocity at mid point is V . \[ \begin{aligned} & y^2=u^2+2 a s \\ & y^2=y^2+2 a s \\ & y^2-v^2=u^2-v^2 \\ & 2 v^2=u^2+v^2 \\ & y^2=\frac{1}{2}\left(u^2+v^2\right) \\ & V=\sqrt{\frac{1}{2}\left(u^2+v^2\right)} \end{aligned} \]
**NCERT-XI-I-18**
TEST SERIES (PHYSICS FST)
266262
lonisation potential of hydrogen atom is 13.6 eV . The energy required to remove an electron from the second orbit of hydrogen is (in eV ):
1 3.4
2 1.89
3 10.2
4 8.5
Explanation:
a Energy of electron in \(n^{\text {th }}\) state of hydrogen atom \(E=\frac{-13.6}{\pi^2} e V \) Energy of electron in second orbit \(E_2=\frac{-13.6}{(2)^2}=-3.4 \mathrm{eV} \) Thus 3.4 eVenergyis required to remove an electron from second orbit
**NCERT-XII-II-300**
TEST SERIES (PHYSICS FST)
266263
A p-n photodiode is fabricated from a semiconductor with a band gap of 3.1 eV . It can detect a signal of wavelength :
1 4000 nm
2 6000 nm
3 4000 A
4 6000 A
Explanation:
c \(\mathrm{A}-\frac{12400 \mathrm{eV}}{3.1 \mathrm{eV}}\) A \(=4000 \mathrm{~A}\).
