A galvanometer, having a resistance of $50 \,\Omega$ gives a full scale deflection for a current of $0.05\, A$. The length in meter of a resistance wire of area of cross-section $2.97× 10^{-2} \,cm^2$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5\, A$ current is (Specific resistance of the wire = $5 × {10^{ - 7}}\,\Omega m$)
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In the figure shown a coil of single turn is wound on a sphere of radius $R$ and mass $m.$ The plane of the coil is parallel to the plane and lies in the equatorial plane of the sphere. Current in the coil is $i$. The value of $B$ if the sphere is in equilibrium is
An ion beam of specific charge $5 \times 10^7$ $coulomb/kg$ enter a uniform magnetic field of $4 \times 10^{-2}\, tesla$ with a velocity $2 \times 10^5\, m/s$ perpendicularly. The radius of the circular path of ions in meter will be
In the hydrogen atom, the electron is making $6.6 \times {10^{15}}\,r.p.s.$ If the radius of the orbit is $0.53 \times {10^{ - 10}}\,metre,$ then magnetic field produced at the centre of the orbit is......$Tesla$
When a galvanometer is shunted with a $4\,\Omega$ resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with $2\,\Omega$ wire, the further reduction in the deflection will be (the main current remains same)
A galvanometer of $50\, ohm$ resistance has $25$ divisions. A current of $4 \times 10^{-4}$ ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25\, volts$, it should be connected with a resistance of
A particle of charge $-q$ and mass $m$ enters a uniform magnetic field $\vec{B}$ at $A$ with speed $v_1$ at an angle $\alpha$ and leaves the field at $C$ with speed $v_2$ at an angle $\beta$ as shown. Then