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The main function of a step-up transformer is to

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D

Increase voltage

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Here are the solutions to your questions:

93. The main function of a step-up transformer is to
A step-up transformer increases voltage and decreases current.
The correct option is D. Increase voltage.

94. The rate at which electric charge flows past a point in a circuit is measured in
The rate of flow of electric charge is defined as electric current. The SI unit for electric current is the ampere.
The correct option is C. amperes.

95. The commercial unit of electrical energy is
The commercial unit of electrical energy is the kilowatt-hour (kWh).
The correct option is D. kilowatt hour.

96. A suitable material for the filament of a bulb should have
A bulb filament needs to glow brightly without melting. This requires a high melting point and high resistance to generate sufficient heat and light.
The correct option is C. high melting point and high resistance.

97. A lighting circuit is supplied through a 5A fuse. What is the maximum number of 75W, 240V bulbs that can be fixed in the circuit?

  1. Calculate the total power the fuse can handle.

    The total power PtotalP_{\text{total}} is given by the product of voltage VV and current II.

    Ptotal=V×IP_{\text{total}} = V \times I Ptotal=240 V×5 A=1200 WP_{\text{total}} = 240 \, \text{V} \times 5 \, \text{A} = 1200 \, \text{W}
  2. Calculate the maximum number of bulbs.

    The maximum number of bulbs nn is the total power divided by the power of a single bulb PbulbP_{\text{bulb}}.

    n=PtotalPbulbn = \frac{P_{\text{total}}}{P_{\text{bulb}}} n=1200 W75 W=16n = \frac{1200 \, \text{W}}{75 \, \text{W}} = 16

    The maximum number of bulbs is B. 16.

    98. Which of the following statements about boiling and evaporation is correct?
    Evaporation occurs at the surface of a liquid and can happen at any temperature below the boiling point. Boiling occurs throughout the liquid at a specific boiling temperature.
    The correct option is D. evaporation takes place at a lower temperature than boiling does.

    99. A body weighs 120N in air, and 80N in water. What is the density of the body in kgm⁻³?

  3. Calculate the buoyant force.

    The buoyant force FBF_B is the apparent loss of weight when the body is immersed in water.

    FB=Wair−WwaterF_B = W_{\text{air}} - W_{\text{water}} FB=120 N−80 N=40 NF_B = 120 \, \text{N} - 80 \, \text{N} = 40 \, \text{N}
  4. Calculate the volume of the body.

    The buoyant force is equal to the weight of the displaced fluid (FB=ρwaterVbodygF_B = \rho_{\text{water}} V_{\text{body}} g). We can also relate the density of the body to the buoyant force.
    The density of the body ρbody\rho_{\text{body}} can be found using the formula:

    ρbody=WairFB×ρwater\rho_{\text{body}} = \frac{W_{\text{air}}}{F_B} \times \rho_{\text{water}}

    Assuming the density of water ρwater=1000 kg/m3\rho_{\text{water}} = 1000 \, \text{kg/m}^3.

    ρbody=120 N40 N×1000 kg/m3\rho_{\text{body}} = \frac{120 \, \text{N}}{40 \, \text{N}} \times 1000 \, \text{kg/m}^3 ρbody=3×1000 kg/m3=3000 kg/m3\rho_{\text{body}} = 3 \times 1000 \, \text{kg/m}^3 = 3000 \, \text{kg/m}^3

    The density of the body is D. 3000.

    100. The principle of floatation states that
    The principle of floatation (Archimedes' principle for floating objects) states that a floating body displaces a weight of fluid equal to its own weight.
    The correct option is C. a floating body displaces its own weight of the fluid.

    101. A mass of 400g was heated and immersed in an insulated copper calorimeter containing 200g of water at 16°C. If the temperature of the water and calorimeter of mass 400g goes to 60°C, find the initial temperature of the metal. (Specific heat capacities of aluminium, copper and water are 110J/KgK, 400J/KgK and 4200 J/KgK respectively.)

  5. Identify the given values and specific heat capacities.

    • Mass of metal mm=400 g=0.4 kgm_m = 400 \, \text{g} = 0.4 \, \text{kg}
    • Specific heat capacity of metal cm=110 J/kgKc_m = 110 \, \text{J/kgK} (assuming the metal is aluminium, as it's the first specific heat given)
    • Mass of water mw=200 g=0.2 kgm_w = 200 \, \text{g} = 0.2 \, \text{kg}
    • Specific heat capacity of water cw=4200 J/kgKc_w = 4200 \, \text{J/kgK}
    • Mass of calorimeter mc=400 g=0.4 kgm_c = 400 \, \text{g} = 0.4 \, \text{kg}
    • Specific heat capacity of calorimeter cc=400 J/kgKc_c = 400 \, \text{J/kgK} (for copper)
    • Initial temperature of water and calorimeter Tw,i=Tc,i=16∘CT_{w,i} = T_{c,i} = 16^\circ \text{C}
    • Final temperature of mixture Tf=60∘CT_f = 60^\circ \text{C}
    • Let the initial temperature of the metal be Tm,iT_{m,i}.
  6. Calculate the heat gained by the water.

    Qw=mwcw(Tf−Tw,i)Q_w = m_w c_w (T_f - T_{w,i}) Qw=0.2 kg×4200 J/kgK×(60∘C−16∘C)Q_w = 0.2 \, \text{kg} \times 4200 \, \text{J/kgK} \times (60^\circ \text{C} - 16^\circ \text{C}) Qw=0.2×4200×44=36960 JQ_w = 0.2 \times 4200 \times 44 = 36960 \, \text{J}
  7. Calculate the heat gained by the calorimeter.

    Qc=mccc(Tf−Tc,i)Q_c = m_c c_c (T_f - T_{c,i}) Qc=0.4 kg×400 J/kgK×(60∘C−16∘C)Q_c = 0.4 \, \text{kg} \times 400 \, \text{J/kgK} \times (60^\circ \text{C} - 16^\circ \text{C}) Qc=0.4×400×44=7040 JQ_c = 0.4 \times 400 \times 44 = 7040 \, \text{J}
  8. Apply the principle of calorimetry.

    Heat lost by metal = Heat gained by water + Heat gained by calorimeter.

    Qm=Qw+QcQ_m = Q_w + Q_c mmcm(Tm,i−Tf)=Qw+Qcm_m c_m (T_{m,i} - T_f) = Q_w + Q_c 0.4 kg×110 J/kgK×(Tm,i−60∘C)=36960 J+7040 J0.4 \, \text{kg} \times 110 \, \text{J/kgK} \times (T_{m,i} - 60^\circ \text{C}) = 36960 \, \text{J} + 7040 \, \text{J} 44×(Tm,i−60)=4400044 \times (T_{m,i} - 60) = 44000
  9. Solve for the initial temperature of the metal Tm,iT_{m,i}.

    Tm,i−60=4400044T_{m,i} - 60 = \frac{44000}{44} Tm,i−60=1000T_{m,i} - 60 = 1000 Tm,i=1000+60T_{m,i} = 1000 + 60 Tm,i=1060∘CT_{m,i} = 1060^\circ \text{C}

    The initial temperature of the metal is 1060∘C\boxed{\text{}1060^\circ \text{C}}.

    102.a. What is a transformer?
    A transformer is an electrical device that transfers electrical energy between two or more circuits through electromagnetic induction. It is used to increase or decrease alternating current (AC) voltages.

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