A horizontal straight wire 10 m long extending from east to west is falling with a speed
Question: A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s1, at right angles to the horizontal component of the earths magnetic field, 0.30 104Wb m2. (a)What is the instantaneous value of the emf induced in the wire? (b)What is the direction of the emf? (c)Which end of the wire is at the higher electrical potentia Solution: Length of the wire,l= 10 m Falling speed of the wire,v= 5.0 m/s Magnetic field strength,B= 0.3 104Wb m2 (a)Emf induced in t...
Read More →A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter
Question: A circular coil of radius $8.0 \mathrm{~cm}$ and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad $\mathrm{s}^{-1}$ in a uniform horizontal magnetic field of magnitude $3.0 \times 10^{-2} \mathrm{~T}$. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance $10 \Omega$, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from? So...
Read More →If (i) , then verify that
Question: If (i) $A=\left[\begin{array}{cc}\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha\end{array}\right]$, then verify that $A^{\prime} A=I$ (ii) $A=\left[\begin{array}{cc}\sin \alpha \cos \alpha \\ -\cos \alpha \sin \alpha\end{array}\right]$, then verify that $A^{\prime} A=I$ Solution: (i) $A=\left[\begin{array}{cc}\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha\end{array}\right]$ $\therefore A^{\prime}=\left[\begin{array}{cc}\cos \alpha -\sin \alpha \\ \sin \alpha \cos \alpha\end{a...
Read More →What do you expect the nature of hydrides is, if formed by elements of atomic
Question: What do you expect the nature of hydrides is, if formed by elements of atomic numbers 15, 19, 23 and 44 with dihydrogen? Compare their behaviour towards water. Solution: The elements of atomic numbers 15, 19, 23, and 44 are phosphorus, potassium, vanadium, and ruthenium respectively. 1) Hydride of phosphorus Hydride of nitrogen (PH3) is a covalent molecule. It is an electron-rich hydride owing to the presence of excess electrons as a lone pair on phosphorus. 2) Hydride of potassium Dih...
Read More →For the matrices $A$ and $B$, verify that $(A B)^{prime}=B^{prime} A^{prime}$ where
Question: For the matrices $A$ and $B$, verify that $(A B)^{\prime}=B^{\prime} A^{\prime}$ where (i) $A=\left[\begin{array}{r}1 \\ -4 \\ 3\end{array}\right], B=\left[\begin{array}{lll}-1 2 1\end{array}\right]$ (ii) $A=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right]$,$B=\left[\begin{array}{lll}1 5 7\end{array}\right]$ Solution: (i) $A B=\left[\begin{array}{r}1 \\ -4 \\ 3\end{array}\right]\left[\begin{array}{lll}-1 2 1\end{array}\right]=\left[\begin{array}{rrr}-1 2 1 \\ 4 -8 -4 \\ -3 6 3\end{a...
Read More →A 1.0 m long metallic rod is rotated with an angular frequency of
Question: A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s1about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. Solution: Length of the rod,l= 1 m Angular frequency,= 400 rad/s Magnetic field strength,B= 0.5 T One end of the rod has zero linear ve...
Read More →A rectangular wire loop of sides
Question: A rectangular wire loop of sides $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3$ T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 $\mathrm{cm} \mathrm{s}^{-1}$ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? Solution: Length of the rectangular wire,l= 8 cm = 0.08 m Width...
Read More →Question: A rectangular wire loop of sides $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3$ T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 $\mathrm{cm} \mathrm{s}^{-1}$ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? Solution: Length of the rectangular wire,l= 8 cm = 0.08 m Width...
Read More →How can saline hydrides remove traces of water from organic compounds?
Question: How can saline hydrides remove traces of water from organic compounds? Solution: Saline hydrides areionic in nature. They react with water to form a metal hydroxide along with the liberation of hydrogen gas. The reaction of saline hydrides with water can be represented as: $\mathrm{AH}_{(s)}+\mathrm{H}_{2} \mathrm{O}_{(n)} \longrightarrow \mathrm{AOH}_{(a q)}+\mathrm{H}_{2(g)}$ (where, A = Na, Ca,) When added to an organic solvent, they react with water present in it. Hydrogen escapes ...
Read More →In how many ways can the letters of the word PERMUTATIONS be arranged if the
Question: In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (ii) there are always 4 letters between P and S? Solution: In the word PERMUTATIONS, there are 2 Ts and all the other letters appear only once. (i) If P and S are fixed at the extreme ends (P at the left end and S at the right end), then 10 letters are left. Hence, in this case, required number of arrangements $=\frac{10 !}{2 !}=1814400$ (ii)...
Read More →A long solenoid with 15 turns per cm has a small loop of area
Question: A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? Solution: Number of turns on the solenoid = 15 turns/cm = 1500 turns/m Number of turns per unit length,n= 1500 turns The solenoid has a small loop of area,A= 2.0 cm2= 2 104m2 Current carried by the solenoid changes from 2...
Read More →What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
Question: What is the difference between the terms hydrolysis and hydration? Solution: Hydrolysis is defined as a chemical reaction in which hydrogen and hydroxide ions (H+and OHions) of water molecule react with a compound to form products. For example: $\mathrm{NaH}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NaOH}+\mathrm{H}_{2}$ Hydration is defined as the addition of one or more water molecules to ions or molecules to form hydrated compounds. For example: $\mathrm{CuSO}_{4}+5 \mathrm{...
Read More →If and, then find
Question: If $A^{\prime}=\left[\begin{array}{cc}-2 3 \\ 1 2\end{array}\right]$ and $B=\left[\begin{array}{rr}-1 0 \\ 1 2\end{array}\right]$, then find $(A+2 B)^{\prime}$ Solution: We know that $A=\left(A^{\prime}\right)^{\prime}$ $\therefore A=\left[\begin{array}{rr}-2 1 \\ 3 2\end{array}\right]$\ $\therefore A+2 B=\left[\begin{array}{rr}-2 1 \\ 3 2\end{array}\right]+2\left[\begin{array}{rr}-1 0 \\ 1 2\end{array}\right]=\left[\begin{array}{rr}-2 1 \\ 3 2\end{array}\right]+\left[\begin{array}{rr}...
Read More →Knowing the properties of H2O and D2O, do you think that D2O can be used for drinking purposes?
Question: Knowing the properties of H2O and D2O, do you think that D2O can be used for drinking purposes? Solution: Heavy water (D2O) acts as a moderator, i.e., it slows the rate of a reaction. Due to this property of D2O, it cannot be used for drinking purposes because it will slow down anabolic and catabolic reactions taking place in the body and lead to a casualty....
Read More →If and, then verify that
Question: If $A^{\prime}=\left[\begin{array}{rr}3 4 \\ -1 2 \\ 0 1\end{array}\right]$ and $B=\left[\begin{array}{rrr}-1 2 1 \\ 1 2 3\end{array}\right]$, then verify that (i) $(A+B)^{\prime}=A^{\prime}+B^{\prime}$ (ii) $(A-B)^{\prime}=A^{\prime}-R^{\prime}$ Solution: (i) It is known that $A=\left(A^{\prime}\right)^{\prime}$ Therefore, we have: $A=\left[\begin{array}{rrr}3 -1 0 \\ 4 2 1\end{array}\right]$ $B^{\prime}=\left[\begin{array}{rr}-1 1 \\ 2 2 \\ 1 3\end{array}\right]$ $A+B=\left[\begin{ar...
Read More →What properties of water make it useful as a solvent?
Question: What properties of water make it useful as a solvent? What types of compound can it (i) dissolve, and (ii) hydrolyse? Solution: A high value of dielectric constants (78.39 C2/Nm2) and dipole moment make water a universal solvent. Water is able to dissolve most ionic and covalent compounds. Ionic compounds dissolve in water because of the ion-dipole interaction, whereas covalent compounds form hydrogen bonding and dissolve in water. Water can hydrolyze metallic and non-metallic oxides, ...
Read More →Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19:
Question: Use Lenzs law to determine the direction of induced current in the situations described by Fig. 6.19: (a)A wire of irregular shape turning into a circular shape; (b)A circular loop being deformed into a narrow straight wire. Solution: According to Lenzs law, the direction of the inducedemfis such that it tends to produce a current that opposes the change in the magnetic flux that produced it. (a)When the shape of the wire changes, the flux piercing through the unit surface area increas...
Read More →Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ).
Question: Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ). (a) (b) (c) (d) (e) (f) Solution: The direction of the induced current in a closed loop is given by Lenzs law. The given pairs of figures show the direction ofthe induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively. Using Lenzs rule, the direction of the induced current in the given situations can be predicted as follows:...
Read More →Describe the usefulness of water in biosphere and biological systems.
Question: Describe the usefulness of water in biosphere and biological systems. Solution: Wateris essential for all forms of life. It constitutes around 65% of the human body and 95% of plants. Water plays an important role in the biosphere owing to its high specific heat, thermal conductivity, surface tension, dipole moment, and dielectric constant. The high heat of vapourization and heat of capacity of water helps in moderating the climate and body temperature ofall living beings. It acts as a...
Read More →In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Question: In how many of the distinct permutations of the letters in MISSISSIPPI do the four Is not come together? Solution: In the given word MISSISSIPPI, I appears 4 times, S appears 4 times, P appears 2 times, and M appears just once. Therefore, number of distinct permutations of the letters in the given word $=\frac{11 !}{4 ! 4 ! 2 !}$ $=\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 4 \times 3 \times 2 \times 1 \times 2 \times 1}$ $=\frac{11 \times 10...
Read More →If and, then verify that
Question: If $A=\left[\begin{array}{rrr}-1 2 3 \\ 5 7 9 \\ -2 1 1\end{array}\right]$ and $B=\left[\begin{array}{rrr}-4 1 -5 \\ 1 2 0 \\ 1 3 1\end{array}\right]$, then verify that (i) $(A+B)^{\prime}=A^{\prime}+B^{\prime}$ (ii) $(A-B)^{\prime}=A^{\prime}-B^{\prime}$ Solution: We have: $A^{\prime}=\left[\begin{array}{rrr}-1 5 -2 \\ 2 7 1 \\ 3 9 1\end{array}\right], B^{\prime}=\left[\begin{array}{rrr}-4 1 1 \\ 1 2 3 \\ -5 0 1\end{array}\right]$ (i) $A+B=\left[\begin{array}{rrr}-1 2 3 \\ 5 7 9 \\ -2...
Read More →Is demineralised or distilled water useful for drinking purposes?
Question: Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful? Solution: Water is an important part of life. It contains several dissolved nutrients that are required by human beings, plants, and animals for survival. Demineralised water is free of all soluble minerals. Hence, it is not fit for drinking. It can be made useful only after the addition of desired minerals in specific amounts, which are important for growth....
Read More →What is meant by ‘demineralised’ water and how can it be obtained?
Question: What is meant by demineralised water and how can it be obtained? Solution: Demineralised water is free from all soluble mineral salts. It does not contain any anions or cations. Demineralised water is obtained by passing water successively through a cation exchange (in the H+form) and an anion exchange (in the OHform) resin. During the cation exchange process, H+exchanges for Na+, Mg2+, Ca2+, and other cations present in water. In the anion exchange process, $\mathrm{OH}^{-}$exchanges ...
Read More →How many words, with or without meaning can be made from the letters of the word MONDAY,
Question: How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel? Solution: There are 6 different letters in the word MONDAY. (i) Number of 4-letter words that can be form...
Read More →The magnetic moment vectors
Question: The magnetic moment vectorssand lassociated with the intrinsic spin angular momentumSand orbital angular momentuml, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by: s= (e/m)S, l=(e/2m)l Which of these relations is in accordance with the result expectedclassically? Outline the derivation of the classical result. Solution: The magnetic moment associated with the orbital angular momentum is valid with the classic...
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