The angle of dip increases as we move from ————–
a) Equator to poles b) Poles to equator c) Dip is equal at all places d) None of these Solution:... View Article
a) Equator to poles b) Poles to equator c) Dip is equal at all places d) None of these Solution:... View Article
Solution: The energy of a photon is given by the formula: E=\frac{hc}{\lambda} =\frac{(6.62*10^{-34})(3*10^{8})}{632.8*10^{-9}} = 3.14*10^{-19}J The energy emitted per second... View Article
Solution: Given: Temperature of coil T_{2}=250K Temperature of condenser T_{1}=300K coefficient of performance of refrigerator = \frac{T_{2}}{T_{1}-T_{2}} = \frac{250}{300-250} =... View Article
a) 2R b) R c) R^{2} d) 1/R Solution: b) R. Explanation: Since, C=4\pi \varepsilon _{0}R C\propto R
a) The direction of the frictional force acting on the book is in the same direction as the frictional force... View Article
Solution: Initial momentum of 3m mass = 0 When the object of mass 3m splits into three fragments of equal... View Article
The process of heat transfer in which heat is transferred with actual migration of medium particles is known as convection.... View Article
a) \(\begin{array}{l}\frac{bv}{Rg}\end{array} \) b) \(\begin{array}{l}\frac{vb^{2}}{Rg}\end{array} \) c) \(\begin{array}{l}\frac{bv^{2}}{Rg}\end{array} \) d) \(\begin{array}{l}\frac{Rg^{2}}{bv}\end{array} \) Solution: c) \(\begin{array}{l}\frac{bv^{2}}{Rg}\end{array} \)
Answer: Potential of a charged sphere is calculated using the below formula: \(\begin{array}{l}V=\frac{kQ}{R} \end{array} \). We know that Q =... View Article
Solution: \(\begin{array}{l}T=I\alpha\end{array} \) \(\begin{array}{l}20=12*\alpha\end{array} \) \(\begin{array}{l}\alpha =\frac{20}{12}\end{array} \) \(\begin{array}{l}\omega_{0}=40\end{array} \) \(\begin{array}{l}\omega =\omega_{0}+\alpha t\end{array} \) \(\begin{array}{l}100 =40+\frac{20}{12}t\end{array} \) \(\begin{array}{l}\frac{60*12}{20}=t\end{array} \) t=36... View Article
Solution: \(\begin{array}{l}a=\frac{y}{I_{g}}\end{array} \) \(\begin{array}{l}=\frac{750*10^{-3}}{15*10^{-3}}\end{array} \) \(\begin{array}{l}=50\Omega \end{array} \) \(\begin{array}{l}I_{g}=\frac{S}{S+a}I\end{array} \) \(\begin{array}{l}15*10^{-3}=\frac{S}{S+50}*25\end{array} \) \(\begin{array}{l}=0.03\Omega\end{array} \)
Solution: For closed tube f_{n}=\frac{nv}{4L} L=1.1m, v=330m/s f_{n}=\frac{n*330}{4*1.1} For highest frequency, f_{h}=\frac{6*330}{4*1.1} = 450 Hz For lowest frequency, f_{1}=\frac{1*330}{4*1.1} =... View Article
Solution: (a) The emf induced in the loop abc is zero. Since the components of ab are opposite to of... View Article
Solution: \(\begin{array}{l}\vec{F}. \vec{\Delta r}=(7\hat{i}+4\hat{j}+3\hat{k}). (2\hat{i}+3\hat{j}+4\hat{k})\end{array} \) = 14+12+12=38J
Solution: \(\begin{array}{l}\oint \vec{B } . \vec{dl}= \mu _{0}I\Rightarrow B.2b= \mu _{0}I\end{array} \) \(\begin{array}{l}B=\frac{\mu_{0} I}{2b}\end{array} \) Net field = 2B Magnetic... View Article
Solution: \(\begin{array}{l}\Delta q \frac{\Delta \phi }{R}\end{array} \) \(\begin{array}{l} i\Delta t = \frac{\Delta \phi }{R}\end{array} \) \(\begin{array}{l}\Delta \phi = i(\Delta t)R\end{array}... View Article
a) Remains the same b) Becomes empty c) Rises up d) Goes down Solution: c) Rises up
Solution: Taking moment about point B N_{A}d=W(d-x)=0 N_{A}=\frac{W(d-x)}{d}
a) g b) g/10 c) (9/10) g d) (10/9) g Answer: b) g/10 Solution: Given. R = 9Mg/10 The cabin is... View Article
a) 4 N b) 6 N c) 8 N d) 2 N Answer: d) 2 N Solution: Now if we consider... View Article