It is given that E= 50\\ V R_eq = R_! +R_2 \\implies R_eq = (100+5) \\Omega = 105 \\Omega So, Current I = \\dfrac{50}{105}A \\implies I = 0.48A P.d. across 5\\ \\Omega =I \ imes 5\\Omega \\implies V = 0.48 \ imes 5 \\implies V = 2.4 V Since the current and voltage value fall in the range of ammeter and voltmeter given in option D.So, option D is correct.