भौतिक विज्ञान (Physics) NEET प्रश्न हिंदी में — उत्तर सहित MCQ अभ्यास

$ l^2 ={ RV \over \rho}$

$ = 1.0 mW = 10^{-3} W $ $ \lambda = 632.8 nm = 632 \times 10^{-9} m $ $\lambda = 632.8 nm = 632 \times 10^{-9} m $ $ P = { nhc \over \lambda } $ $ \therefore n = { p \lambda \over hc } $ $ = { 10^{-3} \times 6.3328 \times 10^{-7} \over 6.625 \times 10^{-34} \times 3 \times 10^ {3}} $ $ ={ 6.328 \times 10^{-20} \over 19.875 \times 10^{26} } $ $=0.318 \times 10^{16} $ $\therefore n = 318 \times 10^{15} s^{-1} $

$ \lambda_1 = 3500 A ^\circ , \lambda _2 =290 nm $ $ h = 6.625 \times 10^ { -24} J.s $ $ { 1 \over 2 } mv^2 max = { hc \lambda } - \phi = eV_0 $ $ \therefore V_01 e = { hc \over \lambda _1 } -\phi $ $ \therefore V_02e = { hc \over \lambda_1} -\phi $ $ \therefore V_02 e = { hc \over \lambda_2} - \phi $ $ (V_02 -V_01) e = hc ( {1\over \lambda_2 } - { 1 \over \lambda_1} )$ $ \therefore V_02 -V_01 = { hc \over e } [ { \lambda_1 - \lambda_2 \over \lambda_1 \lambda_2} ] = 12.42 = [ { 0.6 \over 3.5 \times 2.9 } ]$ $ = 0.7342$ $ = 73.42 \times 10^{-2} V $

$ P = { h \over \lambda } $ $ \therefore P \alpha { 1 \over \lambda }$ $ \therefore { P_1 \over P_2 } = { \lambda_1 \over \lambda_2 } = { 8000 \over 4000} = { 2 \over 1 } $ $ \therefore {P_1 \over P_2} = 2:1 $

$E = K + U $ $ E = { p ^ 2 \over 2m } - eV_{(x)} $ $ \therefore p = ( 2mE + eV_{(x)} )^{1/2} $ $ \therefore \lambda = { h \over p } = { h \over [ 2m ( E + eV_{(x)} )]^{1/2} }$ $ \mu = { \lambda_0 \over \lambda } = { h \over \sqrt { 2mE} } \times { [2m ( E + eV_{(x)})]^{1/2} \over h }$ $ \mu = [ { E + eV_{(x)} \over E} ] ^ { 1 /2} $

$\lambda_1 = 4000 A^\circ = 4 \times 10^ -7 m $ $ \lambda_2 = 3000 A^\circ = 3 \times 10 ^ {-7} m$ $ \triangle = ? $ $ E = { hc \over \lambda } -\phi $ $ E_1 = { hc \over \lambda_1} -\phi ...(1) $ $ E_2 = { hc \over \lambda_2} - \phi...(2) $ $ \therefore E_2 -E_1 = hc ({1 \over \lambda_2 } - { 1 \over \lambda_1} )$ $ = 6.625 \times 10^{-34} \times 3 \times 10^{8} [{10^7 \over 3} - {10^7 \over 4}] = 19.956 \times 10^{-19} [{1 \over 12}]$ $ \therefore \triangle E = 1.04 eV$

प्रतिरोधक के सिरों पर अधिकतम विभवांतर ज्ञात करने के लिए, हम सूत्र $V_{max} = I_{rms} imes R imes \sqrt{2}$ का उपयोग करते हैं। दिया गया है $I_{rms} = 10 ext{ A}$ और $R = 12 \\Omega$, तो $V_{max} = 10 imes 12 imes \sqrt{2} = 169.68 ext{ V}$। अतः अधिकतम विभवांतर 169.68 V है।

$ K = vq $ $ { 1 \over 2 } mv^2 = Vq , m^2 v^2 = 2Vqm $ $ \therefore mv = \sqrt { 2Vqm} $ $ \lambda = { h \over mv } = { h \over \sqrt { 2Vqm } } $ $ \therefore \lambda \alpha { 1 \over \sqrt { qm } }$ $ \therefore { \lambda_\alpha \over \lambda_p} = \sqrt { q_p m_p \over q_\alpha m_\alpha } = \sqrt { em_p \over 2e \times 4m_p } = \sqrt { 1 \over 8} $ $ \therefore {\lambda_\alpha \over \lambda_p } = { 1 \over 2 \sqrt 2 } $