1.The measured values of two resistances R1R_1R1 and R2R_2R2 are R1=(100±0.3) ΩR_1 = (100 \pm 0.3)\,\OmegaR1=(100±0.3)Ω and R2=(200±0.4) ΩR_2 = (200 \pm 0.4)\,\OmegaR2=(200±0.4)Ω. ThenNote: More than one option may be correct.a.The value of R1+R2R_1 + R_2R1+R2 is (300±0.7) Ω(300 \pm 0.7)\,\Omega(300±0.7)Ωb.The value of R2−R1R_2 - R_1R2−R1 is (100±0.1) Ω(100 \pm 0.1)\,\Omega(100±0.1)Ωc.The value of R2−R1R_2 - R_1R2−R1 is (100±0.7) Ω(100 \pm 0.7)\,\Omega(100±0.7)Ωd.When connected in parallel, 1/Rp=1/R1+1/R21/R_p = 1/R_1 + 1/R_21/Rp=1/R1+1/R2 gives Rp=(66.7±0.18) ΩR_p = (66.7 \pm 0.18)\,\OmegaRp=(66.7±0.18)ΩLogin to continueOnly logged in users canattempt or see the solution.